Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 4 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Maths Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 182:

Answer:

i) (x2  x + 3) is a quadratic polynomial. x2  x + 3 = 0 is a quadratic equation.ii) Clearly, (2x2 + 52x  3) is a quadratic polynomial. 2x2 + 52x  3 = 0 is a quadratic equation.iii) Clearly, (2x2 + 7x + 52) is a quadratic polynomial. 2x2 + 7x + 52 = 0 is a quadratic equation.iv) Clearly, (13x2 + 15x  2) is a quadratic polynomial. 13x2 + 15x  2 = 0 is a quadratic equation.v) (x2  3x  x + 4) contains a term with x, i.e, x12, where 12 is not a integer. Therefore, it is not a quadratic polynomial. x2  3x  x + 4 = 0 is not a quadratic equation.vi) x  6x = 3 x2  6 = 3x x2  3x  6 = 0(x23x6) is a quadratic polynomial; therefore, the givenequation is quadratic. vii) x + 2x = x2 x2 + 2 = x3 x3  x2  2 = 0(x3  x2  2) is  not a quadratic polynomial. x3  x2  2 = 0 is not a quadratic equation.viii) x2  1x2 = 5 x4  1 = 5x2 x4  5x2  1 = 0(x4  5x2  1) is a polynomial with degree 4. x4  5x2  1 = 0 is not a quadratic equation.
(ix) x+23=x3-8
x3+6x2+12x+8=x3-86x2+12x+16=0
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) 2x+33x+2=6x-1x-2
6x2+4x+9x+6=6x2-3x+26x2+13x+6=6x2-18x+1231x-6=0
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) x+1x2=2x+1x+3
x2+1x2=2x2+1x+3x2+12=2xx2+1+3x2x4+2x2+1=2x3+2x+3x2x4-2x3-x2-2x+1=0
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

Page No 182:

Answer:

The given equation is (3x2 + 2x  1 = 0).(i) x = (1) L.H.S. = x2 + 2x  1= 3 × (1)2 + 2 × (1)  1= 3  2  1= 0= R.H.S.Thus, (1) is a root of (3x2 + 2x  1 = 0).(ii) On substituting x = 13 in the given equation, we get:L.H.S. = 3x2 + 2x  1  = 3 × 132 + 2 × 13  1 = 3 1× 193+ 23  1= 1 + 2  33 = 03= 0= R.H.S.Thus,  13 is a root of (3x2 + 2x  1 = 0).(iii) On substituting x = 12 in the given equation, we get:L.H.S. = 3x2 + 2x  1  = 3 × 122 + 21 × 121  1 = 3 × 14 1 1= 34  2= 3  84= 54  0Thus, L.H.S.= R.H.S.Hence, 12 is a not solution of  (3x2 + 2x  1= 0).

Page No 182:

Answer:

(i)
It is given that (x=1) is a root of (x2 + kx + 3 = 0). Therefore, (x=1) must satisfy the equation. (1)2 + k × 1 + 3 = 0  k + 4 = 0  k = 4Hence, the required value of k is 4.
So, the equation becomes x2-4x+3=0
On factorising we get;
x2-x-3x+3=0x(x-1)-3(x-1)=0(x-1)(x-3)=0x-1=0 or x-3=0x=1 or x=3
Hence, the other root is 3.

(ii)
It is given that 34 is a root of ax2 + bx  6 = 0; therefore, we have:a × (34)2 + b × 34 6 = 0 9a16 + 3b4 = 6 9a + 12b16 = 6 9a + 12b  96 = 0 3a + 4b = 32        ...(i) Again, (2) is a root of ax2 + bx  6 = 0; therefore, we have:a×(2)2 + b×(2)  6 = 0 4a  2b = 6 2a  b = 3        ...(ii)On multiplying (ii) by 4 and adding the result with (i), we get: 3a + 4b + 8a  4b = 32 + 12 11a = 44 a = 4Putting the value of a in  (ii), we get:2×4  b = 3  b = 3 b = 5Hence, the required values of a and b are 4 and 5, respectively.



 

Page No 182:

Answer:

LHS;
Consider the quadratic equation;
ad2axb+2cdx+bc2=0
Put x=-bcad in the given equation.
ad2a-bcadb+2cd-bcad+bc2=ad2bdad-bcad+2bc-bcad+bc2=ad2bd-bc+2bc-bcad+bc2=ad2bd(bc)-bcad+bc2=-ab2c2d2abd2+bc2=-bc2+bc2=0=RHS
Hence, x=-bcad is a solution to the given quadratic equation.

Page No 182:

Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

x32 or x = -13

Hence, the roots of the given equation are 32 and -13.

Page No 182:

Answer:

4x2 + 5x = 0

x(4x + 5) = 0

x = 0 or 4x + 5 = 0

x = 0 or x = -54

Hence, the roots of the given equation are 0 and -54.

Page No 182:

Answer:

Given:3x2  243 = 0 3(x2  81) = 0 (x)2  (9)2 = 0 (x + 9)(x  9) = 0 x + 9 = 0 or x  9 = 0 x = 9 or  x= 9Hence,9 and 9 are the roots of the equation 3x2243=0.

Page No 182:

Answer:

We write, x=4x-3x as 2x2×-6=-12x2=4x×-3x

 2x2+x-6=02x2+4x-3x-6=02xx+2-3x+2=0x+22x-3=0

x+2=0 or 2x-3=0x=-2 or x=32
Hence, the roots of the given equation are -2 and 32.

Page No 182:

Answer:

We write, 6x=x+5x as x2×5=5x2=x×5x

 x2+6x+5=0x2+x+5x+5=0xx+1+5x+1=0x+1x+5=0

x+1=0 or x+5=0x=-1 or x=-5

Hence, the roots of the given equation are −1 and −5.

Page No 182:

Answer:

We write, -3x=3x-6x as 9x2×-2=-18x2=3x×-6x

 9x2-3x-2=09x2+3x-6x-2=03x3x+1-23x+1=03x+13x-2=0
3x+1=0 or 3x-2=0x=-13 or x=23
Hence, the roots of the given equation are -13 and 23.

Page No 182:

Answer:

Given:x2 + 12x + 35 = 0 x2 + 7x + 5x + 35 = 0 x(x + 7) + 5(x + 7) = 0 (x + 5)(x + 7) = 0 x + 5 = 0 or x + 7 = 0 x = 5 or x = 7Hence,5 and 7 are the roots of the equation x2 + 12x + 35 = 0.

Page No 182:

Answer:

Given:x2 = 18x  77 x2  18x + 77 = 0  x2  (11x + 7x) + 77 = 0 x2  11x  7x + 77 = 0 x(x  11)  7(x  11) = 0 (x  7)(x  11) = 0 x  7 = 0 or x  11 = 0 x = 7 or x = 11Hence, 7 and 11 are the roots of the equation x2 = 18x  77.



Page No 183:

Answer:

Given:6x2 + 11x + 3 = 0 6x2 + 9x + 2x + 3 = 0 3x(2x + 3) + 1(2x + 3) = 0 (3x + 1)(2x + 3) = 0 3x + 1 = 0 or 2x + 3 = 0 x = 13 or x = 32Hence, 13 and 32are the roots of the equation 6x2 + 11x + 3 = 0.

Page No 183:

Answer:

Given:6x2 + x  12 = 0 6x2 + 9x  8x  12 = 0 3x(2x + 3)  4(2x + 3) = 0 (3x  4)(2x + 3) = 0 3x  4 = 0 or 2x + 3 = 0 x = 43 or x = 32Hence, 43 and 32 are the roots of the equation 6x2 + x  12 = 0.

Page No 183:

Answer:

We write, -2x=-3x+x as 3x2×-1=-3x2=-3x×x

 3x2-2x-1=03x2-3x+x-1=03xx-1+1x-1=0x-13x+1=0
x-1=0 or 3x+1=0x=1 or x=-13
Hence, the roots of the given equation are 1 and -13.

Page No 183:

Answer:

Given:4x2  9x = 100 4x2  9x  100 = 0 4x2  (25x  16x)  100 = 0 4x2  25x + 16x  100 = 0 x(4x  25) + 4(4x  25) = 0 (4x  25)(x + 4) = 0 4x  25 = 0 or x + 4 = 0 x = 254 or x = 4Hence, the roots of the equation are 254 and 4.

Page No 183:

Answer:

Given:15x2  28 = x 15x2  x  28 = 0 15x2 (21x  20x)  28 = 0 15x2  21x + 20x  28 = 0 3x(5x  7) + 4(5x  7) = 0 (3x + 4)(5x  7) = 0 3x + 4 = 0 or 5x  7 = 0 x = 43 or x = 75Hence, the roots of the equation are 43 and 75.

Page No 183:

Answer:

Given: 11x = 3x2  3x2 + 11x  4 = 0 3x2 + 12x  x  4 = 0 3x(x + 4)  1(x + 4) = 0 (x + 4)(3x  1) = 0 x + 4 = 0 or 3x  1 = 0 x = 4 or x = 13Hence, the roots of the equation are 4 and 13.

Page No 183:

Answer:

Given:48x2  13x  1 = 0 48x2  (16x  3x)  1 = 0 48x2  16x + 3x  1 = 0 16x(3x  1) + 1(3x  1) = 0 (16x + 1)(3x  1) = 0 16x + 1 = 0 or 3x  1 = 0 x = 116 or x = 13Hence, the roots of the equation are 116 and 13.

Page No 183:

Answer:

We write, 22x=32x-2x as x2×-6=-6x2=32x×-2x

 x2+22x-6=0x2+32x-2x-6=0xx+32-2x+32=0x+32x-2=0
x+32=0 or x-2=0x=-32 or x=2
Hence, the roots of the given equation are -32 and 2.

Page No 183:

Answer:

Consider 3x2+10x-83=0
Factorising by splitting the middle term;
3x2+12x-2x-83=03x(x+43)-2(x+43)=0(3x-2)(x+43)=03x-2=0 or x+43=0x=23 or x=-43
Hence, the roots of the given equation are 23  and -43.

Page No 183:

Answer:

Given:3x2 + 11x + 63 = 0 3x2 + 9x + 2x + 63 = 0 3x(x + 33) + 2(x + 33) = 0 (x + 33)(3x + 2) = 0 x + 33 = 0 or 3x + 2 = 0 x = 33 or x = 23 = 2 × 33 × 3 = 233Hence, the roots of the equation are 33 and 233.

Page No 183:

Answer:

Given:37x2 + 4x  7 = 0 37x2 + 7x  3x  7 = 0 7x(3x + 7)  1(3x + 7) = 0 (3x + 7)(7x  1) = 0 3x + 7 = 0 or 7x  1 = 0 x = 73 or x = 17 = 1 × 77 × 7 = 77Hence, the roots of the equation are 73 and 77.

Page No 183:

Answer:

We write, -6x=7x-13x as 7x2×-137=-91x2=7x×-13x

7x2-6x-137=07x2+7x-13x-137=07xx+7-13x+7=0x+77x-13=0
x+7=0 or 7x-13=0x=-7 or x=137=1377
Hence, the roots of the given equation are -7 and 1377.

Page No 183:

Answer:

Given:46x2  13x  26 = 0 46x2  16x + 3x  26 = 0 42x(3x  22) + 3(3x  22) = 0 (42x + 3)(3x  22) = 0 42x + 3 = 0 or 3x  22 = 0 x = 342 = 3 × 242 × 2 68 or x = 223 = 22 × 33 × 3 = 263Hence, the roots of the equation are 68 and 263.

Page No 183:

Answer:

We write, -26x=-6x-6x as 3x2×2=6x2=-6x×-6x

3x2-26x+2=03x2-6x-6x+2=03x3x-2-23x-2=03x-23x-2=0
3x-22=03x-2=0x=23 =63
Hence, 63 is the repreated root of the given equation.

Page No 183:

Answer:

We write, -22x=-32x+2x as 3x2×-23=-6x2=-32x×2x

3x2-22x-23=03x2-32x+2x-23=03xx-6+2x-6=0x-63x+2=0
x-6=0 or 3x+2=0x=6 or x=-23 =-63
Hence, the roots of the given equation are 6 and -63.

Page No 183:

Answer:

We write, -35x=-25x-5x as x2×10=10x2=-25x×-5x

x2-35x+10=0x2-25x-5x+10=0xx-25-5x-25=0x-25x-5=0
x-5=0 or x-25=0x=5 or x=25
Hence, the roots of the given equation are 5 and 25.

Page No 183:

Answer:

x2-3+1x+3=0x2-3x-x+3=0xx-3-1x-3=0x-3x-1=0
x-3=0 or x-1=0x=3 or x=1
Hence, 1 and 3 are the roots of the given equation.

Page No 183:

Answer:

We write, 33x=53x-23x as x2×-30=-30x2=53x×-23x

x2+33x-30=0x2+53x-23x-30=0xx+53-23x+53=0x+53x-23=0
x+53=0 or x-23=0x=-53 or x=23
Hence, the roots of the given equation are -53 and 23.

Page No 183:

Answer:

We write, 7x=5x+2x as 2x2×52=10x2=5x×2x

2x2+7x+52=02x2+5x+2x+52=0x2x+5+22x+5=02x+5x+2=0
x+2=0 or 2x+5=0x=-2 or x=-52=-522
Hence, the roots of the given equation are -2 and -522.

Page No 183:

Answer:

We write, 13x = 5x + 8x as 5x2×8=40x2=5x×8x
 5x2+13x+8=05x2+5x+8x+8=05xx+1+8x+1=0x+15x+8=0
x+1=0 or 5x+8=0x=-1 or x=-85
Hence, -1 and -85 are the roots of the given equation.

Page No 183:

Answer:

x2-1+2x+2=0x2-x-2x+2=0x(x-1)-2(x-1)=0(x-2)(x-1)=0x-2=0 and x-1=0x=2 and x=1

Page No 183:

Answer:

Given:9x2 + 6x + 1 = 0 9x2 + 3x + 3x + 1 = 0 3x(3x + 1) + 1(3x + 1) = 0 (3x + 1)(3x + 1) = 0 3x + 1 = 0 or 3x + 1 = 0 x = 13 or x = 13Hence, 13is the root of the equation 9x2 + 6x + 1 = 0.

Page No 183:

Answer:

We write, -20x=-10x-10x as 100x2×1=100x2=-10x×-10x
 100x2-20x+1=0100x2-10x-10x+1=010x10x-1-110x-1=010x-110x-1=0
10x-12=010x-1=0x=110
Hence, 110 is the repreated root of the given equation.

Page No 183:

Answer:

We write, -x=-x2-x2 as 2x2×18=x24=-x2×-x2
 2x2-x+18=02x2-x2-x2+18=02xx-14-12x-14=0x-142x-12=0
x-14=0 or 2x-12=0x=14 or x=14
Hence, 14 is the repeated root of the given equation.

Page No 183:

Answer:

Given:10x  1x = 3 10x2  1 = 3x    [Multiplying both sides by x] 10x2  3x  1 = 0 10x2  (5x  2x)  1 = 0 10x2  5x + 2x  1 = 0 5x(2x  1) + 1(2x  1) = 0 (2x  1)(5x + 1) = 0 2x  1 = 0 or 5x + 1 = 0 x = 12 or x = 15Hence, the roots of the equation are 12 and 15.

Page No 183:

Answer:

Given:2x2  5x + 2 = 0 2  5x + 2x2 = 0    [Multiplying both side by x2] 2x2  5x + 2 = 0 2x2  (4x + x) + 2 = 0 2x2  4x  x + 2 = 0 2x(x  2)  1(x  2) = 0 (2x  1)(x  2) = 0 2x  1 = 0 or x  2 = 0 x = 12 or x = 2Hence, the roots of the equation are 12 and 2.

Page No 183:

Answer:

We write, ax=2ax-ax as 2x2×-a2=-2a2x2=2ax×-ax
 2x2+ax-a2=02x2+2ax-ax-a2=02xx+a-ax+a=0x+a2x-a=0
x+a=0 or 2x-a=0x=-a or x=a2
Hence, -a and a2 are the roots of the given equation.

Page No 183:

Answer:

We write, 4bx=2a+bx-2a-bx as 4x2×-a2-b2=-4a2-b2x2=2a+bx×-2a-bx
 4x2+4bx-a2-b2=04x2+2a+bx-2a-bx-a-ba+b=02x2x+a+b-a-b2x+a+b=02x+a+b2x-a-b=0
2x+a+b=0 or 2x-a-b=0x=-a+b2 or x=a-b2
Hence, -a+b2 and a-b2 are the roots of the given equation.

Page No 183:

Answer:

We write, -4a2x=-2a2+b2x-2a2-b2x as 4x2×a4-b4=4a4-b4x2=-2a2+b2x×-2a2-b2x
 4x2-4a2x+a4-b4=04x2-2a2+b2x-2a2-b2x+a2-b2a2+b2=02x2x-a2+b2-a2-b22x-a2+b2=02x-a2+b22x-a2-b2=0
2x-a2+b2=0 or 2x-a2-b2=0x=a2+b22 or x=a2-b22
Hence, a2+b22 and a2-b22 are the roots of the given equation.

Page No 183:

Answer:

We write, 5x=a+3x-a-2x as x2×-a2+a-6=-a2+a-6x2=a+3x×-a-2x
 x2+5x-a2+a-6=0x2+a+3x-a-2x-a+3a-2=0xx+a+3-a-2x+a+3=0x+a+3x-a-2=0
x+a+3=0 or x-a-2=0x=-a+3 or x=a-2
Hence, -a+3 and a-2 are the roots of the given equation.

Page No 183:

Answer:

We write, -2ax=2b-ax-2b+ax as x2×-4b2-a2=-4b2-a2x2=2b-ax×-2b+ax
 x2-2ax-4b2-a2=0x2+2b-ax-2b+ax-2b-a2b+a=0xx+2b-a-2b+ax+2b-a=0x+2b-ax-2b+a=0
x+2b-a=0 or x-2b+a=0x=-2b-a or x=2b+ax=a-2b or x=a+2b
Hence, a-2b and a+2b are the roots of the given equation.

Page No 183:

Answer:

We write, -2b-1x=-b-5x-b+4x as x2×b2-b-20=b2-b-20x2=-b-5x×-b+4x
 x2-2b-1x+b2-b-20=0x2-b-5x-b+4x+b-5b+4=0xx-b-5-b+4x-b-5=0x-b-5x-b+4=0
x-b-5=0 or x-b+4=0x=b-5 or x=b+4
Hence, b-5 and b+4 are the roots of the given equation.

Page No 183:

Answer:

We write, 6x=a+4x-a-2x as x2×-a2+2a-8=-a2+2a-8x2=a+4x×-a-2x
 x2+6x-a2+2a-8=0x2+a+4x-a-2x-a+4a-2=0xx+a+4-a-2x+a+4=0x+a+4x-a-2=0
x+a+4=0 or x-a-2=0x=-a+4 or x=a-2
Hence, -a+4 and a-2 are the roots of the given equation.

Page No 183:

Answer:

Given:abx2 + (b2  ac)x  bc = 0 abx2 + b2x  acx  bc = 0 bx(ax + b)  c(ax + b) = 0 (bx  c)(ax + b) = 0 bx  c = 0 or ax + b = 0 x = cb or x = baHence, the roots of the equation are cb and ba.

Page No 183:

Answer:

We write, -4ax=-b+2ax+b-2ax as x2×-b2+4a2=-b2+4a2x2=-b+2ax×b-2ax
 x2-4ax-b2+4a2=0x2-b+2ax+b-2ax-b-2ab+2a=0xx-b+2a+b-2ax-b+2a=0x-b+2ax+b-2a=0
x-b+2a=0 or x+b-2a=0x=2a+b or x=-b-2ax=2a+b or x=2a-b
Hence, 2a+b and 2a-b are the roots of the given equation.

Page No 183:

Answer:

Given:4x2  2(a2 + b2)x + a2b2 = 0 4x2  2a2x  2b2x + a2b2 = 0 2x(2x  a2)  b2(2x  a2) = 0 (2x  b2)(2x  a2) = 0 2x  b2 = 0 or 2x  a2 = 0 x = b22 or x = a22Hence, the roots of the equation are b22 and a22.

Page No 183:

Answer:

Given:12abx2  (9a2  8b2)x  6ab = 0 12abx2  9a2x + 8b2x  6ab = 0 3ax(4bx  3a) + 2b(4bx  3a) = 0 (3ax + 2b)(4bx  3a) = 0 3ax + 2b = 0 or 4bx  3a = 0 x = 2b3a or x = 3a4bHence, the roots of the equation are 2b3a and 3a4b.

Page No 183:

Answer:

Given:a2b2x2 + b2x  a2x  1 = 0 b2x(a2x + 1)  1(a2x + 1) = 0 (b2x  1)(a2x + 1) = 0 (b2x  1) = 0 or  (a2x + 1) = 0 x = 1b2 or x = 1a2Hence, 1b2 and 1a2 are the roots of the given equation.

Page No 183:

Answer:

We write, -9a+bx=-32a+bx-3a+2bx as 9x2×2a2+5ab+2b2=92a2+5ab+2b2x2=-32a+bx×-3a+2bx
9x2-9a+bx+2a2+5ab+2b2=09x2-32a+bx-3a+2bx+2a+ba+2b=03x3x-2a+b-a+2b3x-2a+b=03x-2a+b3x-a+2b=0
3x-2a+b=0 or 3x-a+2b=0x=2a+b3 or x=a+2b3
Hence, 2a+b3 and a+2b3 are the roots of the given equation.

Page No 183:

Answer:

16x-1=15x+1, x0, -116x-15x+1=116x+16-15xxx+1=1x+16x2+x=1
x2+x=x+16                     Cross multiplicationx2-16=0x+4x-4=0x+4=0 or x-4=0
x=-4 or x=4
Hence, −4 and 4 are the roots of the given equation.

Page No 183:

Answer:

4x-3=52x+3,   x0, -324x-52x+3=38x+12-5xx2x+3=33x+122x2+3x=3
x+42x2+3x=12x2+3x=x+4                 Cross multiplication2x2+2x-4=0x2+x-2=0
x2+2x-x-2=0xx+2-1x+2=0x+2x-1=0x+2=0 or x-1=0
x=-2 or x=1
Hence, −2 and 1 are the roots of the given equation.

Page No 183:

Answer:

3x+1-12=23x-1,    x-1, 133x+1-23x-1=129x-3-2x-2x+13x-1=127x-53x2+2x-1=12
3x2+2x-1=14x-10                    Cross multiplication3x2-12x+9=0x2-4x+3=0x2-3x-x+3=0
xx-3-1x-3=0x-3x-1=0x-3=0 or x-1=0x=3 or x=1
Hence, 1 and 3 are the roots of the given equation.



Page No 184:

Answer:

(i)
1x-1-1x+5=67,   x1, -5x+5-x+1x-1x+5=676x2+4x-5=67x2+4x-5=7
x2+4x-12=0x2+6x-2x-12=0xx+6-2x+6=0x+6x-2=0
x+6=0 or x-2=0x=-6 or x=2
Hence, −6 and 2 are the roots of the given equation.

(ii)
12x-3+1x-5=11912x-3+1x-5=109(x-5)+(2x-3)(2x-3)(x-5)=1093x-82x2-3x-10x+15=1093x-82x2-13x+15=10927x-72=20x2-130x+15020x2-157x+222=020x2-120x-37x+222=020x(x-6)-37(x-6)=0(20x-37)(x-6)=020x-37=0 or x-6=0x=3720 or x=6

Page No 184:

Answer:

12a+b+2x=12a+1b+12x12a+b+2x-12x=12a+1b2x-2a-b-2x2x2a+b+2x=2a+b2ab-2a+b4x2+4ax+2bx=2a+b2ab
4x2+4ax+2bx=-2ab4x2+4ax+2bx+2ab=04xx+a+2bx+a=0x+a4x+2b=0
x+a=0 or 4x+2b=0x=-a or x=-b2
Hence, -a and -b2 are the roots of the given equation.

Page No 184:

Answer:

Given:x+3x-2-1-xx=414(x + 3)(x  2)  (1  x)x = 174 x(x + 3)  (1  x)(x  2)(x  2)x = 174 x2 + 3x  (x  2  x2 + 2x)x2  2x = 174 x2 + 3x + x2  3x + 2x2  2x = 174 2x2 + 2x2  2x = 174 8x2 + 8 = 17x2  34x        [On cross multiplying] 9x2 + 34x + 8 = 0 9x2  34x  8 = 0 9x2  36x + 2x  8 = 0 9x(x  4) + 2(x  4) = 0 (x  4)(9x + 2) = 0 x  4 = 0 or 9x + 2 = 0 x = 4 or x = 29Hence, the roots of the equation are 4 and 29.

Page No 184:

Answer:

3x-47+73x-4=52,   x433x-42+4973x-4=529x2-24x+16+4921x-28=529x2-24x+6521x-28=52
18x2-48x+130=105x-14018x2-153x+270=02x2-17x+30=02x2-12x-5x+30=0
2xx-6-5x-6=0x-62x-5=0x-6=0 or 2x-5=0x=6 or x=52
Hence, 6 and 52 are the roots of the given equation.

Page No 184:

Answer:

(i)
xx-1+x-1x=414,   x0, 1x2+x-12xx-1=174x2+x2-2x+1x2-x=1742x2-2x+1x2-x=174
8x2-8x+4=17x2-17x9x2-9x-4=09x2-12x+3x-4=03x3x-4+13x-4=0
3x-43x+1=03x-4=0 or 3x+1=0x=43 or x=-13
Hence, 43 and -13 are the roots of the given equation.

(ii)
x-12x+1+2x+1x-1=2(x-1)2+(2x+1)2(2x+1)(x-1)=2(x2+1-2x)+(4x2+1+4x)=2(2x+1)(x-1)5x2+2x+2=2(2x2-x-1)5x2+2x+2=4x2-2x-2x2+4x+4=0x2+2x+2x+4=0x(x+2)+2(x+2)=0(x+2)(x+2)=0(x+2)=0 or (x+2)=0x=-2 or x=-2x=-2

Page No 184:

Answer:

xx+1+x+1x=2415,  x0, -1x2+x+12xx+1=3415x2+x2+2x+1x2+x=34152x2+2x+1x2+x=3415
30x2+30x+15=34x2+34x4x2+4x-15=04x2+10x-6x-15=02x2x+5-32x+5=0
2x+52x-3=02x+5=0 or 2x-3=0x=-52 or x=32
Hence, -52 and 32 are the roots of the given equation.

Page No 184:

Answer:

x-4x-5+x-6x-7=313,  x5, 7x-4x-7+x-5x-6x-5x-7=103x2-11x+28+x2-11x+30x2-12x+35=1032x2-22x+58x2-12x+35=103
x2-11x+29x2-12x+35=533x2-33x+87=5x2-60x+1752x2-27x+88=0
2x2-16x-11x+88=02xx-8-11x-8=0x-82x-11=0
x-8=0 or 2x-11=0x=8 or x=112
Hence, 8 and 112 are the roots of the given equation.

Page No 184:

Answer:

x-1x-2+x-3x-4=313,   x2, 4x-1x-4+x-2x-3x-2x-4=103x2-5x+4+x2-5x+6x2-6x+8=1032x2-10x+10x2-6x+8=103
x2-5x+5x2-6x+8=533x2-15x+15=5x2-30x+402x2-15x+25=02x2-10x-5x+25=0
2xx-5-5x-5=0x-52x-5=0x-5=0 or 2x-5=0x=5 or x=52
Hence, 5 and 52 are the roots of the given equation.

Page No 184:

Answer:

Given:1(x  2) + 2(x  1) = 6x (x  1) + 2(x  2)(x  1)(x  2) = 6x 3x  5x2  3x + 2 = 6x 3x2  5x = 6x2  18x + 12        [On cross multiplying] 3x2  13x + 12 = 0 3x2  (9 + 4)x + 12 = 0 3x2  9x  4x + 12 = 0 3x(x  3)  4(x  3) = 0 (3x  4)(x  3) = 0  3x  4 = 0 or x  3 = 0 x = 43 or x = 3Hence, the roots of the equation are 43 and 3.

Page No 184:

Answer:

(i)

1x+1+2x+2=5x+4,   x-1, -2, -4x+2+2x+2x+1x+2=5x+43x+4x2+3x+2=5x+43x+4x+4=5x2+3x+2
3x2+16x+16=5x2+15x+102x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0
x-22x+3=0x-2=0 or 2x+3=0x=2 or x=-32
Hence, 2 and -32 are the roots of the given equation.

(ii)

1x+1+35x+1=5x+4(5x+1)+(3x+3)(x+1)(5x+1)=5x+4(x+4)(5x+1)+(3x+3)=5(x+1)(5x+1)(x+4)8x+4=25x2+30x+58x2+4x+32x+16=25x2+30x+58x2+36x+16=25x2+30x+517x2-6x-11=017x2-17x+11x-11=017x(x-1)+11(x-1)=0(17x+11)(x-1)=0(17x+11)=0 or (x-1)=0x=-1117 or x=1

Page No 184:

Answer:

33x-12x+3-22x+33x-1=5,   x13, -3233x-12-22x+322x+33x-1=539x2-6x+1-24x2+12x+96x2+7x-3=527x2-18x+3-8x2-24x-186x2+7x-3=5
19x2-42x-156x2+7x-3=519x2-42x-15=30x2+35x-1511x2+77x=011xx+7=0
x=0 or x+7=0x=0 or x=-7
Hence, 0 and −7 are the roots of the given equation.

Page No 184:

Answer:

37x+15x-3-45x-37x+1=11,   x35, -1737x+12-45x-325x-37x+1=11349x2+14x+1-425x2-30x+935x2-16x-3=11147x2+42x+3-100x2+120x-3635x2-16x-3=11
47x2+162x-3335x2-16x-3=1147x2+162x-33=385x2-176x-33338x2-338x=0338xx-1=0
x=0 or x-1=0x=0 or x=1
Hence, 0 and 1 are the roots of the given equation.

Page No 184:

Answer:

Given:4x  32x + 1  102x + 14x  3 = 3Putting 4x  32x + 1 = y, we get:y  10y = 3 y2  10y = 3 y2  10 = 3y [On cross multiplying] y2  3y  10 = 0 y2  (5  2)y  10 = 0 y2  5y + 2y  10 = 0 y(y  5) + 2(y  5) = 0 (y  5)(y + 2) = 0 y  5 = 0 or y + 2 = 0 y = 5 or y = 2Case IIf y = 5, we get:4x  32x + 1 = 5 4x  3 = 5(2x + 1) [On cross multiplying] 4x  3 = 10x + 5 6x = 8 6x = 8 x = 8463 x = 43Case IIIf y = 2, we get:4x  32x + 1 2 4x  3 = 2(2x + 1) 4x  3 = 4x  2 8x = 1 x = 18Hence, the roots of the equation are 43 and 18.

Page No 184:

Answer:

Given:xx + 12  5xx + 1 + 6 = 0Putting xx + 1= y, we get:y2  5y + 6 = 0 y2  5y + 6 = 0 y2  (3 + 2)y + 6 = 0 y2  3y  2y + 6 = 0 y(y  3)  2(y  3) = 0 (y  3)(y  2) = 0 y  3 = 0 or y - 2 = 0 y = 3 or y = 2Case IIf y=3, we get:xx + 1 = 3 x = 3(x + 1) [On cross multiplying] x = 3x + 3 x = 32Case IIIf y = 2, we get:xx + 1 = 2 x = 2(x + 1) x = 2x + 2 x = 2 x = 2 Hence, the roots of the equation are 32 and 2.

Page No 184:

Answer:

a(x  b) + b(x  a) = 2 [a(x  b)  1] + [b(x  a)  1] = 0 a  (x  b)x  b + b  (x  a)x  a = 0 a  x + bx  b + a  x + bx  a = 0 (a  x + b)[1(x  b) + 1(x  a)] = 0 (a  x + b)[(x  a) + (x  b)(x  b)(x  a)] = 0 (a  x + b)[2x  (a + b)(x  b)(x  a)] = 0 (a  x + b)[2x  (a + b)] = 0 a  x + b = 0 or 2x  (a + b) = 0 x = a + b or x = a + b2Hence, the roots of the equation are (a + b) and (a + b2).  

Page No 184:

Answer:

a(ax  1) + b(bx  1) = (a + b) [a(ax  1)  b] + [b(bx  1)  a] = 0 a  b(ax  1)ax  1 + b  a(bx  1)bx  1 = 0 a  abx + bax  1 + a  abx + bbx  1 = 0 (a  abx + b)[1(ax  1) + 1(bx  1)] = 0 (a  abx + b)[(bx  1) + (ax  1)(ax  1)(bx  1)] = 0 (a  abx + b)[(a + b)x  2(ax  1)(bx  1)] = 0 (a  abx + b)[(a + b)x  2] = 0 a  abx + b = 0 or (a + b)x  2 = 0 x = (a + b)ab or x = 2(a + b) Hence, the roots of the equation are (a + b)ab and 2(a + b).

Page No 184:

Answer:

 3(x+2) + 3x = 103x.9  + 13x = 10Let 3x be equal to y. 9y + 1y = 10 9y2 + 1 = 10y 9y2 - 10y + 1 = 0 (y  1)(9y  1) = 0 y  1 = 0 or 9y  1 = 0 y = 1 or y = 19 3x = 1 or 3x = 19 3x = 30  or 3x = 32 x = 0 or x = 2Hence, 0 and 2 are the roots of the given equation.

Page No 184:

Answer:

Given:4(x+1) + 4(1x) = 10 4x.4 + 41.14x = 10Let 4xbe y. 4y + 4y = 10 4y2  10y + 4 = 0 4y2  8y  2y + 4 = 0 4y(y  2)  2(y  2) = 0 y = 2 or y = 24 = 12 4x = 2 or 12 4x = 22x = 21  or  22x = 2-1 x = 12 or  x = -12Hence, 12 and -12 are roots of the given equation.

Page No 184:

Answer:

Given:22x  3.2(x+2) + 32 = 0 (2x)2  3.2x.22 + 32 = 0Let 2x be y. y2  12y + 32 = 0 y2  8y  4y + 32 = 0 y(y  8)  4(y  8) = 0 (y  8) = 0 or (y  4) = 0 y = 8 or y = 4 2x = 8 or 2x = 4 2x = 23 or 2x = 22 x= 2 or 3 Hence, 2 and 3 are the roots of the given equation.



Page No 193:

Answer:

(i) 2x2  7x + 6 = 0Here,a = 2,b = 7,c = 6Discriminant D is diven by:D = b2  4ac= (7)2  4 × 2 × 6= 49  48= 1

(ii) 3x2  2x + 8 = 0Here,a = 3,b = 2,c = 8Discriminant D is given by:D = b2  4ac(2)2  4 × 3 × 84  96= 92

(iii) 2x2  52x + 4 = 0Here,a = 2,b = 52,c = 4Discriminant D is given by:D = b2  4ac= (52)2  4 × 2 × 4= (25 × 2)  32= 50  32= 18

(iv)3x2 + 22x  23 = 0Here,a = 3,b = 22,c = 23Discriminant D is given by:D = b2  4ac= (22)2  4 × 3 × (23)= (4 × 2) + (8 × 3)= 8 + 24= 32

(v) x-12x-1=0

2x2-3x+1=0

Comparing it with ax2+bx+c=0, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D = b2-4ac=-32-4×2×1=9-8=1

(vi) 1  x = 2x2 2x2 + x  1 = 0Here,a = 2,b = 1,c = 1Discriminant D is given by:D = b2  4ac= 12  4 × 2(1)= 1 + 8= 9

Page No 193:

Answer:

Given:x2  4x  1 = 0On comparing it with ax2 + bx + c = 0, we get:a = 1, b = 4 and c = 1Discriminant D is given by:D = (b2  4ac)(4)2  4 × 1 × (1)= 16 + 420= 20 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (4) + 202 × 1 = 4 + 252 = 2(2 + 5)2 = (2 + 5)β = b  D2a = (4)  202 = 4  252 = 2(2  5)2 = (2  5)Thus, the roots of the equation are (2 + 5)  and (2  5).

Page No 193:

Answer:

Given:x2  6x + 4 = 0On comparing it with ax2 + bx + c = 0, we get:a = 1, b = 6 and c = 4Discriminant D is given by:D = (b2  4ac)(6)2  4 × 1 × 4= 36  16= 20 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (6) + 202 × 1 = 6 + 252 = 2(3 + 5)2 (3 + 5)β = b  D2a =(6)  202 × 1 = 6  252 = 2(3  5)2 = (3  5)Thus, the roots of the equation are (3 + 25) and (3  25).

Page No 193:

Answer:

The given equation is 2x2+x-4=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = b2-4ac=12-4×2×-4=1+32=33>0

So, the given equation has real roots.

Now, D=33

 α=-b+D2a=-1+332×2=-1+334β=-b-D2a=-1-332×2=-1-334

Hence, -1+334 and -1-334 are the roots of the given equation.

Page No 193:

Answer:

Given:25x2 + 30x + 7 = 0On comparing it with ax2 + bx + c = 0, we get:a = 25, b = 30 and c = 7Discriminant D is given by:D = (b2  4ac)= 302  4 × 25 × 7= 900  700= 200= 200 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = 30 + 2002 × 25 = 30 + 10250 = 10(3 + 2)50 = (3 + 2)5β = b  D2a = 30  2002 × 25 = 30  10250 = 10(3  2)50 = (3  2)5Thus, the roots of the equation are (3 + 2)5 and (3  2)5.

Page No 193:

Answer:

Given:16x2 = 24x + 1⇒ 16x2  24x  1 = 0On comparing it with ax2 + bx + c = 0a = 16, b = 24 and c = 1Discriminant D is given by:D = (b2  4ac)(24)2  4 × 16 × (1)= 576 + (64)= 640 > 0Hence, the roots of the equation are real,Roots α and β are given by:α = b + D2a (24) + 6402 × 16 = 24 + 81032 = 8(3 + 10)32 = (3 + 10)4β = b  D2a = (24)  6402 × 16 = 24  81032 = 8(3  10)32 = (3  10)4Thus, the roots of the equation are (3 + 10)4 and  (3  10)4.

Page No 193:

Answer:

Given:15x2  28 = x⇒ 15x2  x  28 = 0On comparing it with ax2 + bx + c = 0, we get:a = 15, b = 1 and c = 28Discriminant D is given by:D = (b2  4ac)(1)2  4 × 15 × (28)= 1  (1680)= 1 + 1680= 1681= 1681 > 0Hence, the roots of the equation are real.Roots α and β are given by:α = b + D2a = (1) + 16812 × 15 = 1 + 4130 = 4230 = 75β = b  D2a = (1)  16812 × 15 = 1  4130 = 4030 = 43Thus, the roots of the equation are 75 and 43.

Page No 193:

Answer:

The given equation is 2x2-22x+1=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b-22 and c = 1

∴ Discriminant, D = b2-4ac=-222-4×2×1=8-8=0

So, the given equation has real roots.

Now, D=0

 α=-b+D2a=--22+02×2=224=22β=-b-D2a=--22-02×2=224=22

Hence, 22 is the repeated root of the given equation.

Page No 193:

Answer:

The given equation is 2x2+7x+52=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b = 7 and c = 52

∴ Discriminant, D = b2-4ac=72-4×2×52=49-40=9>0

So, the given equation has real roots.

Now, D=9=3

 α=-b+D2a=-7+32×2=-422=-2β=-b-D2a=-7-32×2=-1022=-522

Hence, -2 and -522 are the roots of the given equation.

Page No 193:

Answer:

Given:3x2 + 10x  83 = 0On comparing it with ax2 + bx + c = 0, we get:a = 3, b = 10 and c = 83Discriminant D is given by:D = (b2  4ac)= (10)2  4 × 3 × (83)= 100 + 96= 196 > 0Hence, the roots of the equation are real.Roots α and β are given by: α b + D2a = 10 + 19623 = 10 + 1423 = 423 = 23 = 23 × 33 = 233β = b  D2a = 10  19623 = 10  1423 = 2423 = 123 = 123 × 33 = 1233 = 43Thus, the roots of the equation are 233 and 43.

Page No 193:

Answer:

The given equation is 3x2-22x-23=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b-22 and c = -23

∴ Discriminant, D = b2-4ac=-222-4×3×-23=8+24=32>0

So, the given equation has real roots.

Now, D=32=42

 α=-b+D2a=--22+422×3=6223=6β=-b-D2a=--22-422×3=-2223=-63

Hence, 6 and -63 are the roots of the given equation.

Page No 193:

Answer:

The given equation is 2x2+63x-60=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b63 and c = -60

∴ Discriminant, D = b2-4ac=632-4×2×-60=108+480=588>0

So, the given equation has real roots.

Now, D=588=143

 α=-b+D2a=-63+1432×2=834=23β=-b-D2a=-63-1432×2=-2034=-53

Hence, 23 and -53 are the roots of the given equation.

Page No 193:

Answer:

The given equation is 43x2+5x-23=0.

Comparing it with ax2+bx+c=0, we get

a = 43, b = 5 and c = -23

∴ Discriminant, D = b2-4ac=52-4×43×-23=25+96=121>0

So, the given equation has real roots.

Now, D=121=11

 α=-b+D2a=-5+112×43=683=34β=-b-D2a=-5-112×43=-1683=-233

Hence, 34 and -233 are the roots of the given equation.

Page No 193:

Answer:

The given equation is 3x2-26x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b-26 and c = 2

∴ Discriminant, D = b2-4ac=-262-4×3×2=24-24=0

So, the given equation has real roots.

Now, D=0

 α=-b+D2a=--26+02×3=266=63β=-b-D2a=--26-02×3=266=63

Hence, 63 is the repeated root of the given equation.

Page No 193:

Answer:

The given equation is 23x2-5x+3=0.

Comparing it with ax2+bx+c=0, we get

a = 23, b-5 and c = 3

∴ Discriminant, D = b2-4ac=-52-4×23×3=25-24=1>0

So, the given equation has real roots.

Now, D=1=1

 α=-b+D2a=--5+12×23=643=32β=-b-D2a=--5-12×23=443=33

Hence, 32 and 33 are the roots of the given equation.

Page No 193:

Answer:

The given equation is x2+x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = b2-4ac=12-4×1×2=1-8=-7<0

Hence, the given equation has no real roots (or real roots does not exist).

Page No 193:

Answer:

The given equation is 2x2+ax-a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 2, B = a and C = -a2

∴ Discriminant, D = B2-4AC=a2-4×2×-a2=a2+8a2=9a20

So, the given equation has real roots.

Now, D=9a2=3a

 α=-B+D2A=-a+3a2×2=2a4=a2β=-B-D2A=-a-3a2×2=-4a4=-a

Hence, a2 and -a are the roots of the given equation.

Page No 193:

Answer:

The given equation is x2-3+1x+3=0.

Comparing it with ax2+bx+c=0, we get

a = 1, b-3+1 and c = 3

∴ Discriminant, D = b2-4ac=-3+12-4×1×3=3+1+23-43=3-23+1=3-12>0

So, the given equation has real roots.

Now, D=3-12=3-1

 α=-b+D2a=--3+1+3-12×1=3+1+3-12=232=3β=-b-D2a=--3+1-3-12×1=3+1-3+12=22=1

Hence, 3 and 1 are the roots of the given equation.

Page No 193:

Answer:

The given equation is 2x2+53x+6=0.

Comparing it with ax2+bx+c=0, we get

a = 2, b53 and c = 6

∴ Discriminant, D = b2-4ac=532-4×2×6=75-48=27>0

So, the given equation has real roots.

Now, D=27=33

 α=-b+D2a=-53+332×2=-234=-32β=-b-D2a=-53-332×2=-834=-23

Hence, -32 and -23 are the roots of the given equation.

Page No 193:

Answer:

The given equation is 3x2-2x+2=0.

Comparing it with ax2+bx+c=0, we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = b2-4ac=-22-4×3×2=4-24=-20<0

Hence, the given equation has no real roots (or real roots does not exist).

Page No 193:

Answer:

The given equation is

x+1x=3,  x0x2+1x=3x2+1=3xx2-3x+1=0

This equation is of the form ax2+bx+c=0, where a = 1, b = −3 and c = 1.

∴ Discriminant, D = b2-4ac=-32-4×1×1=9-4=5>0

So, the given equation has real roots.

Now, D=5

 α=-b+D2a=--3+52×1=3+52β=-b-D2a=--3-52×1=3-52

Hence, 3+52 and 3-52 are the roots of the given equation.

Page No 193:

Answer:

The given equation is

1x-1x-2=3,  x0, 2x-2-xxx-2=3-2x2-2x=3-2=3x2-6x

3x2-6x+2=0

This equation is of the form ax2+bx+c=0, where a = 3, b = −6 and c = 2.

∴ Discriminant, D = b2-4ac=-62-4×3×2=36-24=12>0

So, the given equation has real roots.

Now, D=12=23

 α=-b+D2a=--6+232×3=6+236=3+33β=-b-D2a=--6-232×3=6-236=3-33

Hence, 3+33 and 3-33 are the roots of the given equation.

Page No 193:

Answer:

The given equation is

x-1x=3,  x0x2-1x=3x2-1=3xx2-3x-1=0

This equation is of the form ax2+bx+c=0, where a = 1, b = −3 and c = −1.

∴ Discriminant, D = b2-4ac=-32-4×1×-1=9+4=13>0

So, the given equation has real roots.

Now, D=13

 α=-b+D2a=--3+132×1=3+132β=-b-D2a=--3-132×1=3-132

Hence, 3+132 and 3-132 are the roots of the given equation.

Page No 193:

Answer:

The given equation is

mnx2+nm=1-2xm2x2+n2mn=1-2xm2x2+n2=mn-2mnxm2x2+2mnx+n2-mn=0

This equation is of the form ax2+bx+c=0, where a = m2, b = 2mn and c = n2-mn.

∴ Discriminant, D = b2-4ac=2mn2-4×m2×n2-mn=4m2n2-4m2n2+4m3n=4m3n>0

So, the given equation has real roots.

Now, D=4m3n=2mmn

 α=-b+D2a=-2mn+2mmn2×m2=2m-n+mn2m2=-n+mnmβ=-b-D2a=-2mn-2mmn2×m2=-2mn+mn2m2=-n-mnm

Hence, -n+mnm and -n-mnm are the roots of the given equation.

Page No 193:

Answer:

The given equation is 36x2-12ax+a2-b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 36, B-12a and C = a2-b2

∴ Discriminant, D = B2-4AC=-12a2-4×36×a2-b2=144a2-144a2+144b2=144b2>0

So, the given equation has real roots.

Now, D=144b2=12b

 α=-B+D2A=--12a+12b2×36=12a+b72=a+b6β=-B-D2A=--12a-12b2×36=12a-b72=a-b6

Hence, a+b6 and a-b6 are the roots of the given equation.

Page No 193:

Answer:

Given: x2  2ax + (a2  b2) = 0On comparing it with Ax2 + Bx + C = 0, we get:A = 1, B = 2a and C = (a2  b2)Discriminant D is given by: D B2  4AC    (2a)2  4 × 1 × (a2  b2)= 4a2  4a2 + 4b2  = 4b2 > 0Hence, the roots of the equation are real.Roots α and β are given by:α b + D2a = (2a) + 4b22 × 1 = 2a + 2b2 = 2(a + b)2 = (a + b)β b  D2a = (2a)  4b22 × 1 = 2a  2b2 = 2(a  b)2 = (a  b)Hence, the roots of the equation are (a + b) and (ab).

Page No 193:

Answer:

The given equation is x2-2ax-4b2-a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B-2a and C = -4b2-a2

∴ Discriminant, D = B2-4AC=-2a2-4×1×-4b2-a2=4a2+16b2-4a2=16b2>0

So, the given equation has real roots.

Now, D=16b2=4b

 α=-B+D2A=--2a+4b2×1=2a+2b2=a+2bβ=-B-D2A=--2a-4b2×1=2a-2b2=a-2b

Hence, a+2b and a-2b are the roots of the given equation.



Page No 194:

Answer:

The given equation is x2+6x-a2+2a-8=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = 6 and C = -a2+2a-8

∴ Discriminant, D = B2-4AC=62-4×1×-a2+2a-8=36+4a2+8a-32=4a2+8a+4=4a2+2a+1=4a+12>0

So, the given equation has real roots.

Now, D=4a+12=2a+1

 α=-B+D2A=-6+2a+12×1=2a-42=a-2β=-B-D2A=-6-2a+12×1=-2a-82=-a-4=-a+4

Hence, a-2 and -a+4 are the roots of the given equation.

Page No 194:

Answer:

The given equation is x2+5x-a2+a-6=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = 5 and C = -a2+a-6

∴ Discriminant, D = B2-4AC=52-4×1×-a2+a-6=25+4a2+4a-24=4a2+4a+1=2a+12>0

So, the given equation has real roots.

Now, D=2a+12=2a+1

 α=-B+D2A=-5+2a+12×1=2a-42=a-2β=-B-D2A=-5-2a+12×1=-2a-62=-a-3=-a+3

Hence, a-2 and -a+3 are the roots of the given equation.

Page No 194:

Answer:

The given equation is x2-4ax-b2+4a2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = −4a and C = -b2+4a2

∴ Discriminant, D = B2-4AC=-4a2-4×1×-b2+4a2=16a2+4b2-16a2=4b2>0

So, the given equation has real roots.

Now, D=4b2=2b

 α=-B+D2A=--4a+2b2×1=4a+2b2=2a+bβ=-B-D2A=--4a-2b2×1=4a-2b2=2a-b

Hence, 2a+b and 2a-b are the roots of the given equation.

Page No 194:

Answer:

The given equation is 4x2-4a2x+a4-b4=0.

Comparing it with Ax2+Bx+C=0, we get

A = 4, B = −4a2 and C = a4-b4

∴ Discriminant, D = B2-4AC=-4a22-4×4×a4-b4=16a4-16a4+16b4=16b4>0

So, the given equation has real roots.

Now, D=16b4=4b2

 α=-B+D2A=--4a2+4b22×4=4a2+b28=a2+b22β=-B-D2A=--4a2-4b22×4=4a2-b28=a2-b22

Hence, 12a2+b2 and 12a2-b2 are the roots of the given equation.

Page No 194:

Answer:

The given equation is 4x2+4bx-a2-b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = 4, B = 4b and C = -a2-b2

∴ Discriminant, D = B2-4AC=4b2-4×4×-a2-b2=16b2+16a2-16b2=16a2>0

So, the given equation has real roots.

Now, D=16a2=4a

 α=-B+D2A=-4b+4a2×4=4a-b8=a-b2β=-B-D2A=-4b-4a2×4=-4a+b8=-a+b2

Hence, 12a-b and -12a+b are the roots of the given equation.

Page No 194:

Answer:

The given equation is x2-2b-1x+b2-b-20=0.

Comparing it with Ax2+Bx+C=0, we get

A = 1, B = -2b-1 and C = b2-b-20

∴ Discriminant, D = B2-4AC=-2b-12-4×1×b2-b-20=4b2-4b+1-4b2+4b+80=81>0

So, the given equation has real roots.

Now, D=81=9

 α=-B+D2A=--2b-1+92×1=2b+82=b+4β=-B-D2A=--2b-1-92×1=2b-102=b-5

Hence, b+4 and b-5 are the roots of the given equation.

Page No 194:

Answer:

Given:3a2x2 + 8abx + 4b2 = 0   On comparing it with Ax2 + Bx + C = 0, we get:A = 3a2, B = 8ab and C = 4b2   Discriminant D is given by:D = (B2  4AC)             = (8ab)2  4 × 3a2 × 4b2     = 16a2b2 > 0Hence, the roots of the equation are real.Roots α and β are given by:α b + D2a = 8ab + 16a2b22 × 3a2 = 8ab + 4ab6a2 = 4ab6a2 = 2b3aβ = b  D2a = 8ab  16a2b22 × 3a2 = 8ab  4ab6a2 = 12ab6a2 = 2baThus, the roots of the equation are 2b3a and 2ba.

Page No 194:

Answer:

The given equation is a2b2x2-4b4-3a4x-12a2b2=0.

Comparing it with Ax2+Bx+C=0, we get

A = a2b2, B = -4b4-3a4 and C = -12a2b2

∴ Discriminant, D = B2-4AC=-4b4-3a42-4×a2b2×-12a2b2=16b8-24a4b4+9a8+48a4b4=16b8+24a4b4+9a8=4b4+3a42>0

So, the given equation has real roots.

Now, D=4b4+3a42=4b4+3a4

 α=-B+D2A=--4b4-3a4+4b4+3a42×a2b2=8b42a2b2=4b2a2β=-B-D2A==--4b4-3a4-4b4+3a42×a2b2=-6a42a2b2=-3a2b2

Hence, 4b2a2 and -3a2b2 are the roots of the given equation.

Page No 194:

Answer:

Given:12abx2  (9a2  8b2)x  6ab = 0     On comparing it with Ax2 + Bx + C = 0, we get:     = 12ab, B = (9a2  8b2) and C = 6abDiscriminant D is given by: D B2  4AC  = [(9a2  8b2)]2  4 × 12ab × (6ab)     = 81a4  144a2b2 + 64b4 + 288a2b2   = 81a4 + 144a2b2 + 64b4   = (9a2 + 8b2)2 > 0Hence, the roots of the equation are equal.Roots α and β are given by:α B + D2A = [(9a2  8b2)] + (9a2 + 8b2)22 × 12ab = 9a2  8b2 + 9a2 + 8b224ab = 18a224ab = 3a4bβ B  D2A = [(9a2  8b2)]  (9a2+8b2)22 × 12ab = 9a2  8b2  9a2  8b224ab = 16b224ab = 2b3aThus, the roots of the equation are 3a4b and 2b3a.



Page No 201:

Answer:

(i) The given equation is 2x2-8x+5=0.

This is of the form ax2+bx+c=0, where a = 2, b = −8 and c = 5.

∴ Discriminant, D = b2-4ac=-82-4×2×5=64-40=24>0

Hence, the given equation has real and unequal roots.

(ii) The given equation is 3x2-26x+2=0.

This is of the form ax2+bx+c=0, where a = 3, b-26 and c = 2.

∴ Discriminant, D = b2-4ac=-262-4×3×2=24-24=0

Hence, the given equation has real and equal roots.

(iii) The given equation is 5x2-4x+1=0.

This is of the form ax2+bx+c=0, where a = 5, b = −4 and c = 1.

∴ Discriminant, D = b2-4ac=-42-4×5×1=16-20=-4<0

Hence, the given equation has no real roots.

(iv) The given equation is

5xx-2+6=05x2-10x+6=0

This is of the form ax2+bx+c=0, where a = 5, b = −10 and c = 6.

∴ Discriminant, D = b2-4ac=-102-4×5×6=100-120=-20<0

Hence, the given equation has no real roots.

(v) The given equation is 12x2-415x+5=0.

This is of the form ax2+bx+c=0, where a = 12, b-415 and c = 5.

∴ Discriminant, D = b2-4ac=-4152-4×12×5=240-240=0

Hence, the given equation has real and equal roots.

(vi) The given equation is x2-x+2=0.

This is of the form ax2+bx+c=0, where a = 1, b = −1 and c = 2.

∴ Discriminant, D = b2-4ac=-12-4×1×2=1-8=-7<0

Hence, the given equation has no real roots.

Page No 201:

Answer:

The given equation is 2a2+b2x2+2a+bx+1=0.
D=2a+b2-4×2a2+b2×1       =4a2+2ab+b2-8a2+b2       =4a2+8ab+4b2-8a2-8b2       =-4a2+8ab-4b2       =-4a2-2ab+b2       =-4a-b2<0
Hence, the given equation has no real roots.



Page No 202:

Answer:

Given: x2 + px  q2 = 0Here, a = 1, b = p and c = q2Discriminant D is given by:D = (b2  4ac)= p2  4 × 1 × (q2)= (p2 + 4q2) > 0D>0 for all real values of p and q.Thus, the roots of the equation are real.

Page No 202:

Answer:

Given: 3x2 + 2kx + 27 = 0Here, a = 3, b = 2k and c = 27It is given that the roots of the equation are real and equal; therefore, we have:D = 0 (2k)2  4 × 3 × 27 = 0 4k2  324 = 0 4k2 = 324 k2 = 81 k = ±9 k = 9 or k = 9

Page No 202:

Answer:

The given equation is

kxx-25+10=0kx2-25kx+10=0

This is of the form ax2+bx+c=0, where a = k, b = -25k and c = 10.

D=b2-4ac=-25k2-4×k×10=20k2-40k

The given equation will have real and equal roots if D = 0.

20k2-40k=020kk-2=0k=0 or k-2=0k=0 or k=2

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

Page No 202:

Answer:

The given equation is 4x2+px+3=0.

This is of the form ax2+bx+c=0, where a = 4, b = p and c = 3.

D=b2-4ac=p2-4×4×3=p2-48

The given equation will have real and equal roots if D = 0.

p2-48=0p2=48p=±48=±43

Hence, 43 and -43 are the required values of p.

Page No 202:

Answer:

The given equation is 9x2-3kx+k=0.

This is of the form ax2+bx+c=0, where a = 9, b = −3k and c = k.

D=b2-4ac=-3k2-4×9×k=9k2-36k

The given equation will have real and equal roots if D = 0.

9k2-36k=09kk-4=0k=0 or k-4=0k=0 or k=4

But, k ≠ 0        (Given)

Hence, the required value of k is 4.

Page No 202:

Answer:

(i)
The given equation is 3k+1x2+2k+1x+1=0.

This is of the form ax2+bx+c=0, where a = 3k +1, b = 2(k + 1) and c = 1.

D=b2-4ac       =2k+12-4×3k+1×1       =4k2+2k+1-43k+1       =4k2+8k+4-12k-4
        =4k2-4k

The given equation will have real and equal roots if D = 0.

4k2-4k=04kk-1=0k=0 or k-1=0k=0 or k=1

Hence, 0 and 1 are the required values of k.

(ii)
x2+k2x+k-1+2=0x2+2kx+k2-k+2=0For real and equal roots D=0D=b2-4ac=0D=2k2-41k2-k+2D=4k2-4k2+4k-8=04k-8=0k=2

Page No 202:

Answer:

The given equation is 2p+1x2-7p+2x+7p-3=0.

This is of the form ax2+bx+c=0, where a = 2p +1, b = (7p + 2) and c = 7p − 3.

D=b2-4ac       =-7p+22-4×2p+1×7p-3       =49p2+28p+4-414p2+p-3       =49p2+28p+4-56p2-4p+12       
       =-7p2+24p+16

The given equation will have real and equal roots if D = 0.

-7p2+24p+16=07p2-24p-16=07p2-28p+4p-16=07pp-4+4p-4=0
p-47p+4=0p-4=0 or 7p+4=0p=4 or p=-47

Hence, 4 and -47 are the required values of p.

Page No 202:

Answer:

The given equation is p+1x2-6p+1x+3p+9=0.

This is of the form ax2+bx+c=0, where a = p +1, b = 6(p + 1) and c = 3(p + 9).

D=b2-4ac       =-6p+12-4×p+1×3p+9       =12p+13p+1-p+9       =12p+12p-6
 
The given equation will have real and equal roots if D = 0.

12p+12p-6=0p+1=0 or 2p-6=0p=-1 or p=3

But, p-1           (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes 4x2-24x+36=0.

4x2-24x+36=04x2-6x+9=0x-32=0x-3=0
x=3

Hence, 3 is the repeated root of this equation.

Page No 202:

Answer:

It is given that −5 is a root of the quadratic equation 2x2+px-15=0.

2-52+p×-5-15=0-5p+35=0p=7

The roots of the equation px2+px+k=0 = 0 are equal.

D=0p2-4pk=072-4×7×k=049-28k=0
k=4928=74

Thus, the value of k is 74.

Page No 202:

Answer:

It is given that 3 is a root of the quadratic equation x2-x+k=0.

32-3+k=0k+6=0k=-6

The roots of the equation x2+2kx+k2+2k+p=0 are equal.

D=02k2-4×1×k2+2k+p=04k2-4k2-8k-4p=0-8k-4p=0
p=8k-4=-2kp=-2×-6=12

Hence, the value of p is 12.

Page No 202:

Answer:

It is given that −4 is a root of the quadratic equation x2+2x+4p=0.

-42+2×-4+4p=016-8+4p=04p+8=0p=-2

The equation x2+px1+3k+73+2k=0 has equal roots.

 D=0p1+3k2-4×1×73+2k=0-21+3k2-283+2k=041+6k+9k2-283+2k=0
41+6k+9k2-21-14k=09k2-8k-20=09k2-18k+10k-20=09kk-2+10k-2=0
k-29k+10=0k-2=0 or 9k+10=0k=2 or k=-109

Hence, the required value of k is 2 or -109.

Page No 202:

Answer:

Given: (1 + m2)x2 + 2mcx + (c2  a2) = 0Here, a = (1 + m2), b = 2mc and c = (c2  a2)It is given that the roots of the equation are equal; therefore, we have:D = 0 (b2  4ac) = 0 (2mc)2  4 × (1 + m2) × (c2  a2) = 0 4m2c2  4(c2  a2 + m2c2  m2a2) = 0 4m2c2  4c2 + 4a2  4m2c2 + 4m2a2 = 0 4c2 + 4a2 + 4m2a2 = 0 a2 + m2a2 = c2 a2(1 + m2) = c2 c2 = a2(1 + m2)Hence proved.

Page No 202:

Answer:

Given: (c2  ab)x2  2(a2  bc)x + (b2  ac) = 0Here, a = (c2  ab), b = 2(a2  bc), c = (b2  ac)It is given that the roots of the equation are real and equal; therefore, we have:D=0(b2  4ac) = 0 {2(a2  bc)}2  4 × (c2  ab) × (b2  ac) = 0 4(a4  2a2bc + b2c2)  4(b2c2  ac3  ab3 + a2bc) = 0 a4  2a2bc + b2c2  b2c2 + ac3 + ab3  a2bc = 0 a4  3a2bc + ac3 + ab3 = 0 a(a3  3abc + c3 + b3) = 0Now,a = 0 or a3  3abc + c3 + b3 = 0a = 0 or a3 + b3 + c3 = 3abc

Page No 202:

Answer:

Given: 2x2 + px + 8 = 0Here, a = 2, b = p and c = 8Discriminant D is given by:D = (b2  4ac)p2  4 × 2 × 8(p2  64)If D  0, the roots of the equation will be real. (p2  64)  0 (p + 8) (p - 8)  0 p  8 and p  -8Thus, the roots of the equation are real for p  8 and p-8.

Page No 202:

Answer:

Given: (α  12)x2 + 2(α  12)x + 2 = 0Here, a = (α  12), b = 2(α  12) and c = 2It is given that the roots of the equation are equal; therefore, we have:D = 0 (b2  4ac) = 0 {2(α  12)}2  4 × (α  12) × 2 = 0 4(α2  24α + 144)  8(α  12) = 0 4α2  96α + 576  8α + 96 = 0 4α2  104α + 672 = 0 α2  26α + 168 = 0 α2  14α  12α + 168 = 0 α(α  14)  12(α  14) = 0 (α  14)(α  12) = 0 α = 14 or α = 12If the value of α is 12, the given equation becomes non-quadratic.Therefore, the valueof α will be 14 for the equation to have equal roots.



Page No 203:

Answer:

Given: 9x2 + 8kx + 16 = 0Here,a = 9, b = 8k and c = 16It is given that the roots of the equation are real and equal; therefore, we have:D = 0 (b2  4ac) = 0 (8k)2  4 × 9 × 16 = 0 64k2  576 = 0 64k2 = 576 k2 = 9 k = ±3 k = 3 or k = 3

Page No 203:

Answer:

(i) The given equation is kx2+6x+1=0.

D=62-4×k×1=36-4k

The given equation has real and distinct roots if D > 0.

36-4k>04k<36k<9

(ii) The given equation is x2-kx+9=0.

D=-k2-4×1×9=k2-36

The given equation has real and distinct roots if D > 0.

k2-36>0k-6k+6>0k<-6 or k>6

(iii) The given equation is 9x2+3kx+4=0.

D=3k2-4×9×4=9k2-144

The given equation has real and distinct roots if D > 0.

9k2-144>09k2-16>0k-4k+4>0k<-4 or k>4

(iv) The given equation is 5x2-kx+1=0.

D=-k2-4×5×1=k2-20

The given equation has real and distinct roots if D > 0.

k2-20>0k2-252>0k-25k+25>0k<-25 or k>25

Page No 203:

Answer:

The given equation is a-bx2+5a+bx-2a-b=0.

D=5a+b2-4×a-b×-2a-b       =25a+b2+8a-b2
Since a and b are real and ab, so a-b2>0 and a+b2>0.

8a-b2>0    .....(1)               (Product of two positive numbers is always positive)      

Also, 25a+b2>0             .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

25a+b2+8a-b2>0                 (Sum of two positive numbers is always positive)

D>0

Hence, the roots of the given equation are real and unequal.

Page No 203:

Answer:

Ans

Page No 203:

Answer:

Ans

Page No 203:

Answer:

It is given that the roots of the equation ax2+2bx+c=0 are real.
D1=2b2-4×a×c04b2-ac0b2-ac0             .....1

Also, the roots of the equation bx2-2acx+b=0 are real.
D2=-2ac2-4×b×b04ac-b20-4b2-ac0
b2-ac0                    .....2

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
b2-ac=0b2=ac

Page No 203:

Answer:

Since,  x=-12 is a root of the quadratic equation 3x2 + 2kx + 3 = 0, 
then, it must satisfies the equation.

3-122+2k-12+3=0314-k+3=034-k+3=03-4k+124=03-4k+12=04k=15k=154Hence, the value of k is 154.

Page No 203:

Answer:

Since, x=-13 is a root of the quadratic equation 2x2 + 2x + k = 0, 
then, it must satisfies the equation.

2-132+2-13+k=0219-23+k=029-23+k=02-6+9k9=0-4+9k=09k=4k=49Hence, the value of k is 49.

Page No 203:

Answer:

Given: x2 – 8x + 18 = 0

x2-8x+18=0Adding and subtracting 12×82, we getx2-8x+18+42-42=0x2-8x+16+18-16=0x-42+2=0x-42=-2x-4=±-2But, -2 is not a real number.

Hence, the quadratic equation x2 – 8x + 18 = 0 has no real solution.

Page No 203:

Answer:

Let 4x2 –12x – k = 0 be a quadratic equation.

It is given that, it has no real roots.

Discriminant<0b2-4ac<0-122-44-k<0144+16k<016k<-144k<-9

Hence, the values of k must be less than –9.

Page No 203:

Answer:

Let one root be α and the other root be 1α.

The given equation is 3x– 10k = 0.

Product of roots = k3
α×1α=k31=k3k=3

Hence, the value of k for which the roots of the equation 3x2 – 10x + k = 0 are reciprocal of each other is 3.

Page No 203:

Answer:

Let one root be α and the other root be 1α.

The given equation is 5x2 +13k = 0.

Product of roots = k5
α×1α=k51=k5k=5

Hence, the value of k is 5.

Page No 203:

Answer:

Let 3x2 + kx + 3 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

Discriminant=0b2-4ac=0k2-433=0k2=36k=±6

Hence, the values of k are –6 and 6.

Page No 203:

Answer:

Let 9x2 – 3ax + 1 = 0 be a quadratic equation.

It is given that, it has real and equal roots.

Discriminant=0b2-4ac=0-3a2-491=09a2=36a2=4a=±2

Hence, the values of a are –2 and 2.

Page No 203:

Answer:

Let x2 + k(2x + – 1) + 2 = 0 be a quadratic equation.

x2 + k(2x + – 1) + 2 = 0
x2 + 2xk + k2 – k + 2 = 0

It is given that, it has real and equal roots.

Discriminant=0b2-4ac=02k2-41k2-k+2=04k2-4k2+4k-8=04k=8k=2

Hence, the value of k is 2.



Page No 224:

Answer:

Let the required natural number be x.

According to the given condition,

x+x2=156x2+x-156=0x2+13x-12x-156=0xx+13-12x+13=0
x+13x-12=0x+13=0 or x-12=0x=-13 or x=12

x = 12     (x cannot be negative)

Hence, the required natural number is 12.

Page No 224:

Answer:

Let the required natural number be x.

According to the given condition,

x+x=132

Putting x=y or x = y2, we get

y2+y=132y2+y-132=0y2+12y-11y-132=0yy+12-11y+12=0
y+12y-11=0y+12=0 or y-11=0y=-12 or y=11

y = 11         (y cannot be negative)

Now,

x=11x=112=121

Hence, the required natural number is 121.

Page No 224:

Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

x28-x=19228x-x2=192x2-28x+192=0x2-16x-12x+192=0
xx-16-12x-16=0x-12x-16=0x-12=0 or x-16=0x=12 or x=16

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

Page No 224:

Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
x2+x+12=365x2+x2+2x+1=3652x2+2x-364=0x2+x-182=0
x2+14x-13x-182=0xx+14-13x+14=0x+14x-13=0x+14=0 or x-13=0
x=-14 or x=13
x = 13                (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

Page No 224:

Answer:

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,
x2+x+22=514x2+x2+4x+4=5142x2+4x-510=0x2+2x-255=0
x2+17x-15x-255=0xx+17-15x+17=0x+17x-15=0x+17=0 or x-15=0
x=-17 or x=15
x = 15             (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.

Page No 224:

Answer:

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,
x2+x+22=452x2+x2+4x+4=4522x2+4x-448=0x2+2x-224=0
x2+16x-14x-224=0xx+16-14x+16=0x+16x-14=0x+16=0 or x-14=0
x=-16 or x=14
x = 14             (x is a positive even number)

When x = 14,
x + 2 = 14 + 2 = 16

Hence, the required numbers are 14 and 16.

Page No 224:

Answer:

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,
xx+1=306x2+x-306=0x2+18x-17x-306=0xx+18-17x+18=0
x+18x-17=0x+18=0 or x-17=0x=-18 or x=17
x = 17           (x is a positive integer)

When x = 17,
x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

Page No 224:

Answer:

 Let the required numbers be x and (x+3).According to the question: x(x + 3) = 504 x2 + 3x = 504 x2 + 3x  504 = 0 x2 + (24  21)x  504 = 0 x2 + 24x  21x  504 = 0 x(x + 24)  21(x + 24) = 0 (x + 24)(x  21) = 0 x + 24 = 0 or x  21 = 0 x = 24 or x =  21If x 24, the numbers are 24 and {(24 + 3) 21}.If x = 21, the numbers are 21 and {(21 + 3) = 24}.Hence, the numbers are (24, 21) and (21, 24).

Page No 224:

Answer:

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
3x×3x+1=6489x2+x=648x2+x=72x2+x-72=0
x2+9x-8x-72=0xx+9-8x+9=0x+9x-8=0x+9=0 or x-8=0
x=-9 or x=8
x = 8           (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

Page No 224:

Answer:

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,
xx+2=483x2+2x-483=0x2+23x-21x-483=0xx+23-21x+23=0
x+23x-21=0x+23=0 or x-21=0x=-23 or x=21
x = 21           (x is a positive odd integer)

When x = 21,
x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23.

Page No 224:

Answer:

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
xx+2=288x2+2x-288=0x2+18x-16x-288=0xx+18-16x+18=0
x+18x-16=0x+18=0 or x-16=0x=-18 or x=16
x = 16           (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

Page No 224:

Answer:

Let the required natural numbers be x and (9 − x).

According to the given condition,
1x+19-x=129-x+xx9-x=1299x-x2=129x-x2=18
x2-9x+18=0x2-6x-3x+18=0xx-6-3x-6=0x-3x-6=0
x-3=0 or x-6=0x=3 or x=6

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.

Page No 224:

Answer:

Let the required natural numbers be x and (15 − x).

According to the given condition,
1x+115-x=31015-x+xx15-x=3101515x-x2=31015x-x2=50
x2-15x+50=0x2-10x-5x+50=0xx-10-5x-10=0x-5x-10=0
x-5=0 or x-10=0x=5 or x=10

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.

Page No 224:

Answer:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
1x>1x+3

According to the given condition,
1x-1x+3=328x+3-xxx+3=3283x2+3x=328x2+3x=28
x2+3x-28=0x2+7x-4x-28=0xx+7-4x+7=0x+7x-4=0
x+7=0 or x-4=0x=-7 or x=4
x = 4             (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.

Page No 224:

Answer:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
1x>1x+5

According to the given condition,
1x-1x+5=514x+5-xxx+5=5145x2+5x=514x2+5x=14
x2+5x-14=0x2+7x-2x-14=0xx+7-2x+7=0x+7x-2=0
x+7=0 or x-2=0x=-7 or x=2
x = 2             (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.



Page No 225:

Answer:

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,
7x2+7x+12=122549x2+49x2+2x+1=122549x2+49x2+98x+49=122598x2+98x-1176=0
x2+x-12=0x2+4x-3x-12=0xx+4-3x+4=0x+4x-3=0
x+4=0 or x-3=0x=-4 or x=3
x = 3            (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.

Page No 225:

Answer:

Let the natural number be x.

According to the given condition,
x+1x=658x2+1x=6588x2+8=65x8x2-65x+8=0
8x2-64x-x+8=08xx-8-1x-8=0x-88x-1=0x-8=0 or 8x-1=0
x=8 or x=18
x = 8         (x is a natural number)

Hence, the required number is 8.

Page No 225:

Answer:

Let the two parts be x and (57 − x).

According to the given condition,
x57-x=68057x-x2=680x2-57x+680=0x2-40x-17x+680=0
xx-40-17x-40=0x-40x-17=0x-40=0 or x-17=0x=40 or x=17

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.

Page No 225:

Answer:

Let the two parts be x and (27 − x).

According to the given condition,
1x+127-x=32027-x+xx27-x=3202727x-x2=32027x-x2=180
x2-27x+180=0x2-15x-12x+180=0xx-15-12x-15=0x-12x-15=0
x-12=0 or x-15=0x=12 or x=15

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.

Page No 225:

Answer:

Let the larger and smaller parts be x and y, respectively.According to the question:x y = 16 ...(i)2x2 = y2 + 164 ...(ii)From (i), we get: x = 16  y ...(iii) From (ii) and (iii), we get:2(16  y)2 = y2 + 164 2(256  32y + y2) = y2 + 164 512  64y + 2y2 = y2 + 164 y2  64y + 348 = 0 y2  (58 + 6)y + 348 = 0 y2  58y  6y + 348 = 0  y(y  58)  6(y  58) = 0 (y  58)(y  6) = 0 y  58 = 0 or y  6 = 0 y = 6   ( y < 16)Putting the value of y in equation (iii), we get:  x = 16  6 = 10Hence, the two natural numbers are 6 and 10.

Page No 225:

Answer:

 Let the two natural numbers be x and  y.According to the question:x2 + y2 = 25(x + y) ...(i)x2 + y2 = 50(x  y)   ...(ii)From (i) and (ii), we get:25(x + y) = 50(x  y) x + y = 2(x  y) x + y = 2x  2y y + 2y = 2x  x 3y = x     ...(iii)  From (ii) and (iii), we get:(3y)2 + y2 = 50(3y  y) 9y2 + y2 = 100y 10y2 = 100y y = 10 From (iii), we have:3 × 10 = x 30 = x Hence, the two natural numbers are 30 and 10.

Page No 225:

Answer:

Let the greater number be x and the smaller number be y.According to the question:x2  y2 = 45          ...(i)y2 = 4x                    ...(ii)From (i) and (ii), we get:x2  4x = 45 x2  4x  45 = 0 x2  (9  5)x  45 = 0 x2  9x + 5x  45 = 0 x(x  9) + 5(x  9) = 0 (x  9)(x + 5) = 0 x  9 = 0 or x + 5 = 0 x = 9 or x = 5 x = 9      ( x is a natural number)Putting the value of x in equation (ii), we get:y2 = 4 × 9 y2 = 36 y = 6Hence, the two numbers are 9 and 6.

Page No 225:

Answer:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,
x2+x+1x+2=46x2+x2+3x+2=462x2+3x-44=02x2+11x-8x-44=0
x2x+11-42x+11=02x+11x-4=02x+11=0 or x-4=0x=-112 or x=4
x = 4           (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

Page No 225:

Answer:

 Let the digits at units and tens places be x and y, respectively. Original number = 10y + x According to the question:10y + x = 4(x + y) 10y + x = 4x + 4y 3x  6y = 0 3x = 6y x = 2y ....(i) Also,10y x = 2xy 10y + 2y = 2.2y.y           [ From (i)] 12y = 4y2 y = 3 From (i), we get:x = 2 × 3 =  6 Original number = 10 × 3 + 6 = 36

Page No 225:

Answer:

​​Let the digits at units and tens places be x and y, respectively. ∴xy=14 y=14x              ...(i) According to the question:(10y+x)+45=10x+y9y9x=45yx=5                                      ...(ii) From (i) and (ii), we get:14xx=514x2x=514x2=5xx25x14=0x2(72)x14=0x27x+2x14=0x(x7)+2(x7)=0(x7)(x+2)=0x7=0 or x+2=0x=7 or x=2x=7   ( the digit cannot be negative) Putting x=7 in equation (i), we get:y=2∴ Required number=10×2+7=27

Page No 225:

Answer:

 Let the numerator be x. Denominator = x + 3 Original number = xx + 3 According to the question:xx + 3 +  1(xx + 3) =  2910 xx + 3 + x + 3x = 2910 x2 + (x + 3)2x(x + 3) = 2910 x2 + x2 + 6x + 9x2 + 3x = 2910 2x2 + 6x + 9x2 + 3x = 2910 29x2 + 87x = 20x2 + 60x + 90 9x2 + 27x  90 = 0 9x2 + 3x - 10 = 0x2 + 3x - 10 = 0x2 + 5x - 2x - 10 = 0xx+5-2x+5 = 0x-2x+5 = 0x-2 = 0  or  x + 5 = 0x=2 or x = -5rejectedSo, numerator = x = 2denominator = x + 3 = 2 + 3 = 5So, required fraction = 25

Page No 225:

Answer:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = x-3x
If 1 is added to the denominator, then the new fraction obtained is x-3x+1.

According to the given condition,
x-3x+1=x-3x-115x-3x-x-3x+1=115x-3x+1-xx-3xx+1=115x2-2x-3-x2+3xx2+x=115
x-3x2+x=115x2+x=15x-45x2-14x+45=0x2-9x-5x+45=0
xx-9-5x-9=0x-5x-9=0x-5=0 or x-9=0x=5 or x=9

When x = 5,
x-3x=5-35=25

When x = 9,
x-3x=9-39=69=23             (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is 25.

Page No 225:

Answer:

Let the required number be x.

According to the given condition,
x+1x=2130x2+1x=613030x2+30=61x30x2-61x+30=0
30x2-36x-25x+30=06x5x-6-55x-6=05x-66x-5=05x-6=0 or 6x-5=0
x=65 or x=56
Hence, the required number is 56 or 65.

Page No 225:

Answer:

Let there be x rows.Then, the number of students in each row will also be x. Total number of students = (x2 + 24)According to the question:(x + 1)2  25 = x2 + 24 x2 + 2x + 1  25  x2  24 =  0 2x  48 = 0 2x = 48 x = 24 Total number of students = 242 + 24 = 576 + 24 = 600

Page No 225:

Answer:

Let the total number of students be x.According to the question:300x  300x + 10 = 1300(x + 10)  300xx(x + 10) = 1 300x + 3000  300xx2 + 10x = 1 3000 = x2 + 10x x2 + 10x  3000 = 0 x2 + (60  50)x  3000 = 0 x2 + 60x  50x  3000 = 0 x(x + 60)  50(x + 60) = 0 (x + 60)(x  50) = 0 x = 50 or x =  60x cannot be negative; therefore, the total number of students is 50.



Page No 226:

Answer:

Let the marks of Kamal in mathematics and english be x and y, respectively.According to the question:x + y = 40         ...(i) Also,(x + 3)(y  4) = 360 (x + 3)(40  x  4) = 360    [From(i)] (x + 3)(36  x) = 360 36x  x2 + 108  3x = 360 33x  x2  252 = 0 x2 + 33x  252 = 0 x2  33x + 252 = 0 x2  (21 + 12)x + 252 = 0 x2  21x  12x + 252 = 0 x(x  21)  12(x  21) = 0 (x  21)(x  12) = 0 x = 21 or x = 12If x = 21, y = 40  21 = 19Thus, Kamal scored 21 and 19 marks in mathematics and english, respectively.If x = 12, y = 40  12 = 28Thus, Kamal scored 12 and 28 marks in mathematics and english, respectively.

Page No 226:

Answer:

Let x be the number of students who planned a picnic.

∴ Original cost of food for each member = ₹2000x

Five students failed to attend the picnic. So, (x − 5) students attended the picnic.

∴ New cost of food for each member = ₹2000x-5

According to the given condition,
2000x-5 − ₹2000x = ₹20
2000x-2000x+10000xx-5=2010000x2-5x=20x2-5x=500x2-5x-500=0
x2-25x+20x-500=0xx-25+20x-25=0x-25x+20=0x-25=0 or x+20=0
x=25 or x=-20
x = 25                (Number of students cannot be negative)

Number of students who attended the picnic = x − 5 = 25 − 5 = 20

Amount paid by each student for the food = ₹200025-5 = ₹200020 = ₹100

Page No 226:

Answer:

Let the original price of the book be ₹x.

∴ Number of books bought at original price for ₹600 = 600x

If the price of a book is reduced by ₹5, then the new price of the book is ₹(x − 5).

∴ Number of books bought at reduced price for ₹600 = 600x-5

According to the given condition,
600x-5-600x=4600x-600x+3000xx-5=43000x2-5x=4x2-5x=750
x2-5x-750=0x2-30x+25x-750=0xx-30+25x-30=0x-30x+25=0
x-30=0 or x+25=0x=30 or x=-25
x = 30               (Price cannot be negative)

Hence, the original price of the book is ₹30.

Page No 226:

Answer:

Let the original duration of the tour be x days.

∴ Original daily expenses = ₹10,800x

If he extends his tour by 4 days, then his new daily expenses = ₹10,800x+4

According to the given condition,

10,800x − ₹10,800x+4 = ₹90

10800x+43200-10800xxx+4=9043200x2+4x=90x2+4x=480x2+4x-480=0
x2+24x-20x-480=0xx+24-20x+24=0x+24x-20=0x+24=0 or x-20=0
x=-24 or x=20
x = 20              (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

Page No 226:

Answer:

Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.

According to the given condition,
x+328-x-4=180x+324-x=180-x2+21x+72=180x2-21x+108=0
x2-12x-9x+108=0xx-12-9x-12=0x-12x-9=0x-12=0 or x-9=0
x=12 or x=9

When x = 12,
28 − x = 28 − 12 = 16

When x = 9,
28 − x = 28 − 9 = 19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

Page No 226:

Answer:

Let number of pens bought = x
Let the price per pen = Rs y
It is given that the price of x pens is Rs 180
xy=180          ...ix=180y
Also, given that if 3 more pens are purchased for the same amount, the price per pen gets reduced by Rs 3
(y-3)(x+3)=180          ...(ii)
Put the value of x in (ii) we get
180x-3(x+3)=180180-3xx(x+3)=180(180-3x)(x+3)=180x180x+540-3x2-9x=180x3x2+9x-540=0x2+3x-180=0x2+15x-12x-180=0x(x+15)-12(x+15)=0(x+15)(x-12)=0x+15=0 or x-12=0x=-15 or x=12
We ignore the negative value. 
x=12
Therefore, number of pens is 12.

Page No 226:

Answer:

Let the cost price of the article be ₹x.

∴ Gain percent = x%

According to the given condition,
x + ₹x100×x = ₹75                            (Cost price + Gain = Selling price)
100x+x2100=75x2+100x=7500x2+100x-7500=0x2+150x-50x-7500=0
xx+150-50x+150=0x-50x+150=0x-50=0 or x+150=0x=50 or x=-150
x = 50                   (Cost price cannot be negative)

Hence, the cost price of the article is ₹50.

Page No 226:

Answer:

(i)
Let the present age of the son be x years.

∴ Present age of the man = x2 years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x2 − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

x2-1=8x-1x2-1=8x-8x2-8x+7=0x2-7x-x+7=0
xx-7-1x-7=0x-1x-7=0x-1=0 or x-7=0x=1 or x=7
x = 7                (Man's age cannot be 1 year) 

Present age of the son = 7 years

Present age of the man = 72 years = 49 years

(ii)
Let the age of man be m and the age of son be s
It is given that man is 312 times as old as his son.
m=312sm=72s          ...i
Also given that 
m2+s2=1325          .....(ii)
Put value of (i) in (ii), we get
72s2+s2=132549s2+4s24=132553s2=5300s2=100s=±10
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
m=72sm=72×10m=35
So, age of man = 35 years
Age of son = 10 years

Page No 226:

Answer:

Let the present age of Meena be x years.

Meena's age 3 years ago = (x − 3) years

Meena's age 5 years hence = (x + 5) years

According to the given condition,
1x-3+1x+5=13x+5+x-3x-3x+5=132x+2x2+2x-15=13x2+2x-15=6x+6
x2-4x-21=0x2-7x+3x-21=0xx-7+3x-7=0x-7x+3=0
x-7=0 or x+3=0x=7 or x=-3
x = 7                 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

Page No 226:

Answer:

Let the present ages of the boy and his brother be x years and (25  x) years.According to the question:x(25  x) = 126 25x  x2 = 126 x2  (18 + 7)x + 126 = 0 x2  18x  7x + 126 = 0 x(x  18)  7(x  18) = 0 (x  18)(x  7) = 0 x  18 = 0 or  x  7 = 0 x = 18 or x = 7 x = 18        ( Present age of the boy cannot be less than his brother)If x=18, we have:Present ages of the boy = 18 yearsPresent age of his brother = (2518) years = 7 yearsThus, the present ages of the boy and his brother are 18 years and 7 years, respectively.

Page No 226:

Answer:

Let the present age of Meena be x years. According to the question:(x  5)(x + 8) = 30 x2 + 3x  40 = 30 x2 + 3x  70 = 0 x2 + (10  7)x  70 = 0 x2 + 10x  7x  70 = 0 x(x + 10)  7(x + 10) = 0 (x + 10)(x  7) = 0 x + 10 = 0 or  x  7 = 0 x = 10 or x = 7 x = 7      (  Age cannot be negative)Thus, the present age of Meena is 7 years.



Page No 227:

Answer:

Let son's age 2 years ago be x years. Then,

Man's age 2 years ago = 3x2 years

∴ Son's present age = (x + 2) years

Man's present age = (3x2 + 2) years

In three years time,

Son's age = (x + 2 + 3) years = (x + 5) years

Man's age = (3x2 + 2 + 3) years = (3x2 + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3x2 + 5 = 4(x + 5)

3x2+5=4x+203x2-4x-15=03x2-9x+5x-15=03xx-3+5x-3=0
x-33x+5=0x-3=0 or 3x+5=0x=3 or x=-53
x = 3              (Age cannot be negative)

Son's present age = (x + 2) years = (3 + 2) years = 5 years

Man's present age = (3x2 + 2) years = (3 × 9 + 2) years = 29 years

Page No 227:

Answer:

Let the first speed of the truck be x km/h.

∴ Time taken to cover 150 km = 150x h                                 Time =DistanceSpeed

New speed of the truck = (x + 20) km/h

∴ Time taken to cover 200 km = 200x+20 h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
150x+200x+20=5150x+3000+200xxx+20=5350x+3000=5x2+20x350x+3000=5x2+100x
5x2-250x-3000=0x2-50x-600=0x2-60x+10x-600=0xx-60+10x-60=0
x-60x+10=0x-60=0 or x+10=0x=60 or x=-10
x = 60                  (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.

Page No 227:

Answer:

Let the original speed of the plane be x km/h.

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = 1500x h                             Time=DistanceSpeed

Time taken to reach the destination at actual speed = 1500x+100 h

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

1500x=1500x+100+12                          30 min=3060h=12h1500x-1500x+100=121500x+150000-1500xxx+100=12150000x2+100x=12
x2+100x=300000x2+100x-300000=0x2+600x-500x-300000=0xx+600-500x+600=0
x+600x-500=0x+600=0 or x-500=0x=-600 or x=500
x = 500                 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

Page No 227:

Answer:

Let the usual speed of the train be x km/h.

∴ Reduced speed of the train = (x − 8) km/h

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = 480x h                  Time=DistanceSpeed

Time taken by the train to cover the distance at reduced speed = 480x-8 h

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

480x-8=480x+3480x-8-480x=3480x-480x+3840xx-8=33840x2-8x=3
x2-8x=1280x2-8x-1280=0x2-40x+32x-1280=0xx-40+32x-40=0
x-40x+32=0x-40=0 or x+32=0x=40 or x=-32
x = 40              (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.

Page No 227:

Answer:

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = 54x h                                 Time =DistanceSpeed

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = 63x+6 h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
54x+63x+6=354x+324+63xxx+6=3117x+324=3x2+6x117x+324=3x2+18x
3x2-99x-324=0x2-33x-108=0x2-36x+3x-108=0xx-36+3x-36=0
x-36x+3=0x-36=0 or x+3=0x=36 or x=-3
x = 36                  (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.

Page No 227:

Answer:

Distance covered by the train = 180 km
We know that distance covered(d)=speed(s)×time(t)
s×t=180s=180t          ...i
Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour. 
(s+9)(t-1)=180          ...(ii)
Put the value of (i) in (ii), we get
180t+9(t-1)=180(180+9t)(t-1)=180t180t-180+9t2-9t=180t9t2-9t-180=0t2-t-20=0t2-5t+4t-20=0t(t-5)+4(t-5)=0(t+4)(t-5)=0(t+4)=0 or (t-5)=0t=-4 or t=5
Ignore the negative value
So, time taken = 5 hours
From (i)
s=180t=1805=36
Hence, the speed is 36 km/h.

Page No 227:

Answer:

Distance covered by the train = 300 km
We know that distance covered(d)=speed(s)×time(t)
s×t=300t=300s          ...(i)
Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours. 
(s+5)(t-2)=300          .....(ii)
Put the value of (i) in (ii), we get
s+5300s-2=300(s+5)300-2ss=300(s+5)(300-2s)=300s300s-2s2+1500-10s=300s-2s2+1500-10s=0-s2+750-5s=0s2+5s-750=0s2+30s-25s-750=0s(s+30)-25(s+30)=0(s-25)(s+30)=0(s-25)=0 or (s+30)=0s=25 or s=-30
Ignore the negative value
Therefore, the original speed = 25 km/h

Page No 227:

Answer:

Let the usual speed be x km/hr.According to the question:300x  300(x + 5) = 2 300(x + 5)  300xx(x + 5) = 2 300x + 1500  300xx2 + 5x = 2 1500 = 2(x2 + 5x) 1500 = 2x2 + 10x x2 + 5x  750 = 0 x2 + (30  25)x  750 = 0 x2 + 30x  25x  750 = 0 x(x + 30)  25(x + 30) = 0 (x + 30)(x  25) = 0 x = 30 or x = 25The usual speed cannot be negative; therefore, the speed is 25 km/hr.

Page No 227:

Answer:

Let the speed of the Deccan Queen be x km/hrAccording to the question:Speed of another train = (x - 20) km/h192x  20  192x = 4860 4x  20  4x = 160 4x  4(x  20)(x  20)x = 160 4x  4x + 80x2  20x = 160 80x2  20x = 160 x2  20x = 4800 x2  20x  4800 = 0 x2  (80  60)x  4800 = 0 x2  80x + 60x  4800 = 0 x(x  80) + 60(x  80) = 0 (x  80)(x + 60) =  0 x = 80 or x = 60 The value of x cannot be negative; therefore, the original speed of Deccan Queen is 80 km/hr.



Page No 228:

Answer:

Let the speed of the stream be x km/hr.Given:Speed of the boat = 18 km/hr Speed downstream = (18 + x) km/hr     Speed upstream = (18  x) km/hr 24(18  x)  24(18 + x) = 1 1(18  x)  1(18 + x) = 124 18 + x  18 + x(18  x)(18 + x) = 124 2x182  x2 = 124 324  x2 = 48x 324  x2  48x = 0 x2 + 48x  324 = 0 x2 + (54  6)x  324 = 0 x2 + 54x  6x  324 = 0 x(x + 54)  6(x + 54) = 0 (x + 54)(x  6) = 0 x = 54 or x = 6The value of x cannot be negative; thereore, the speed of the stream is 6 km/hr.

Page No 228:

Answer:

Let, speed of stream be x km/h
Speed of boat = 15 km/h
Distance from each side = 30 km
We know that time taken =distance coveredSpeed
Total speed of the boat while going upstream =15-x km/h
Time taken to go upstream =3015-x hrs
Total speed of boat while going downstream =15+x km/h
Time taken to go downstream =3015+x hrs
Total time of the journey =412hrs = 4.5 hrs
3015-x+3015+x=4.530(15+x)+(15-x)(15)2-x2=4.53015+x+15-x=4.5(15)2-x23030=4.5225-x230×304.5=225-x2225-x2=200x2=25x=±5
Ignore the negative value. 
So, the speed of the stream = x = 5 km/h

Page No 228:

Answer:

Let the speed of the stream be x km/hr. Downstream speed= (9 + x) km/hrUpstream speed= (9  x) km/hrDistance covered downstream = Distance covered upstream= 15 kmTotal time taken = 3 hours 45 minutes =  3+4560 minutes = 22560 minutes = 154minutes15(9 + x) + 15(9  x) = 154 1(9 + x) + 1(9  x) = 14 9  x + 9 + x(9 + x)(9  x) = 14 1892  x2 = 14 1881  x2 = 14 81  x2 = 72 81  x2  72 = 0 x2 + 9 = 0 x2 = 9 x = 3 or x = 3The value of x cannot be negative; therefore, the speed of the stream is 3 km/hr.

Page No 228:

Answer:

Let B takes x days to complete the work. Therefore, A will take (x  10) days. 1x + 1(x  10) = 112 (x  10) + xx(x  10) = 112 2x  10x2  10x = 112 x2  10x = 12(2x  10) x2  10x = 24x  120 x2  34x + 120 = 0 x2  (30 + 4)x + 120 = 0 x2  30x  4x + 120 = 0 x(x  30)  4(x  30) = 0 (x  30)(x  4) = 0 x = 30 or x = 4Number of days to complete the work by B cannot be less than that by A; therefore, we get:x = 30Thus, B completes the work in 30 days.

Page No 228:

Answer:

Let one tap fills the tank in x hours.Therefore, the other tap will fill the tank in x + 3 hours.Time taken by both taps, running together, to fill the tank= 3113 hours = 4013 hoursPart filled by one tap in 1 hour = 1xPart filled by the other tap in 1 hour = 1x + 3Part filled by both taps, running together, in 1 hour = 1x + 1x + 3 1x + 1x + 3 = 14013 (x + 3) + xx(x + 3) = 1340 2x + 3x2 + 3x = 1340 13x2 + 39x = 80x + 120 13x2  41x  120 = 0 13x2  (65  24)x  120 = 0 13x2  65x + 24x  120 = 0 13x(x  5) + 24(x  5) = 0 (x  5)(13x + 24) = 0 x  5 = 0 or 13x + 24 = 0⇒ x = 5 or x 2413 x = 5    ( time cannot be a negative fraction)Thus, one pipe will take 5 hour and the other will take {(5 + 3) = 8} hours to fill the tank.

Page No 228:

Answer:

Let the time taken by one pipe to fill the tank be x minutes.

∴ Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

∴ Volume of the tank filled by one pipe in 1 minute = Vx

⇒ Volume of the tank filled by one pipe in 1119 minutes = Vx×1119=Vx×1009

Similarly,

Volume of the tank filled by the other pipe in 1119 minutes = Vx+5×1119=Vx+5×1009

Now,

Volume of the tank filled by one pipe in 1119 minutes + Volume of the tank filled by the other pipe in 1119 minutes = V
V1x+1x+5×1009=V1x+1x+5=9100x+5+xxx+5=91002x+5x2+5x=9100
200x+500=9x2+45x9x2-155x-500=09x2-180x+25x-500=09xx-20+25x-20=0
x-209x+25=0x-20=0 or 9x+25=0x=20 or x=-259

x = 20                    (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min

Page No 228:

Answer:

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = Vx

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = Vx×6

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = Vx-9×6

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
V1x+1x-9×6=V1x+1x-9=16x-9+xxx-9=162x-9x2-9x=16
12x-54=x2-9xx2-21x+54=0x2-18x-3x+54=0xx-18-3x-18=0
x-18x-3=0x-18=0 or x-3=0x=18 or x=3


For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

x = 18                   

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

Page No 228:

Answer:

Let the length and breadth of the rectangle be 2x m and x m, respectively.According to the question:2x × x = 288 2x2 = 288 x2 = 144 x = 12 or x 12 x = 12        ( x cannot be negative) Length = 2 × 12 = 24 m     Breadth = 12 m

Page No 228:

Answer:

Let the length and breadth of the rectangle be 3x m and x m, respectively.According to the question:3x × x = 147 3x2 = 147 x2 = 49 x = 7 or x 7 x = 7     (x cannot be negative) Length = 3 × 7 = 21m      Breadth = 7 m

Page No 228:

Answer:

Let the breadth of the rectangular hall be x metre.Therefore, the length of the rectangular hall will be (x + 3)metre.According to the question:x(x + 3) = 238 x2 + 3x = 238 x2 + 3x  238 = 0 x2 + (17  14)x  238 = 0 x2 + 17x  14x  238 = 0 x(x + 17)  14(x + 17) = 0 (x + 17)(x  14) = 0 x = 17 or x = 14But the value x cannot be negative.Therefore, the breadth of the hall is 14 metre and the length is 17 metre.

Page No 228:

Answer:

Given:
Perimeter of a rectangular plot = 60 m
Area = 200 m2

Let the length of rectangular plot be x m.
Perimeter=602Length+Breadth=60Length+Breadth=30Breadth=30-Length

Then, the breadth is (30 − x) m.

According to the question,

Length×Breadth=Areax30-x=20030x-x2=200x2-30x+200=0x2-10x-20x+200=0xx-10-20x-10=0x-10x-20=0x=10, 20For x=10,30-x=20For x=20,30-x=10

Hence, the dimensions of the plot are 10 m and 20 m.

Page No 228:

Answer:

Let the width of the path be x m.  Length of the field including the path = 16 + x + x = 16 + 2xBreadth of the field including the path = 10 + x + x = 10 + 2xNow,(Area of the field including path)  (Area of the field excluding path) = Area of the path (16 + 2x)(10 + 2x)  (16 × 10) = 120 160 + 32x + 20x + 4x2  160 = 120 4x2 + 52x  120 = 0 x2 + 13x  30 = 0 x2 + (15  2)x + 30 = 0 x2 + 15x  2x + 30 = 0 x(x + 15)  2(x + 15) = 0 (x  2)(x + 15) = 0 x  2 = 0 or x + 15 = 0 x = 2 or x = 15 x = 2    ( Width cannot be negative)Thus, the width of the path is 2 m.



Page No 229:

Answer:

Let the length of the side of the first and the second square be x and y, respectively.

According to the question:x2 + y2 = 640             ...(i)Also,4x  4y = 64 x  y = 16 x = 16 + y

Putting the value of x in (i), we get:

x2 + y2 = 640 (16 + y)2 + y2 = 640 256 + 32y + y2 + y2 = 640 2y2 + 32y  384 = 0 y2 + 16y  192 = 0 y2 + (24  8)y  192 = 0 y2 + 24y  8y  192 = 0 y(y + 24)  8(y + 24) = 0 (y + 24)(y  8) = 0 y = 24 or y = 8 y = 8    ( Side cannot be negative) x = 16 + y = 16 + 8 = 24 mThus, the sides of the squares are 8 m and 24 m.

Page No 229:

Answer:

Let the breadth of rectangle be x cm.According to the question:Side of the square = (x + 4) cm Length of the rectangle = {3(x + 4)} cmIt is given that the areas of the rectangle and square are same. 3(x + 4) × x = (x + 4)2 3x2 + 12x = (x + 4)2 3x2 + 12x = x2 + 8x + 16 2x2 + 4x  16 = 0 x2 + 2x  8 = 0 x2 + (4  2)x  8 = 0 x2 + 4x  2x  8 = 0 x(x + 4)  2(x + 4) = 0 (x + 4)(x  2) = 0 x = 4 or x = 2 x = 2    ( The value of x cannot be negative)Thus, the breadth of the rectangle is 2 cm and length is {3(2 + 4) =18} cm. Also, the side of the square is 6 cm.

Page No 229:

Answer:

Let the length and breadth of the rectangular garden be x and y metre, respectively.Given:xy = 180 sq m       ...(i)   and2y + x = 39 x = 39  2yPutting the value of x in (i), we get:(39  2y)y = 180 39y  2y2 = 180 39y  2y2  180 = 0 2y2  39y + 180 = 0 2y2  (24 + 15)y + 180 = 0 2y2  24y  15y + 180 = 0 2y(y  12)  15(y  12) = 0 (y  12)(2y  15) = 0 y = 12 or y = 152 = 7.5If y = 12, x = 39  24 = 15If y = 7.5, x = 39  15 = 24Thus, the length and breadth of the garden are (15 m and 12 m) or (24 m and 7.5 m), respectively.

Page No 229:

Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (+ 10) cm.
Area of triangle = 12 x (x + 10) = 600 x(x + 10) = 1200 x2 + 10x  1200 = 0 x2 + (40  30)x  1200 = 0 x2 + 40x  30x  1200 = 0 x(x + 40)  30(x + 40) = 0 (x + 40)(x  30) = 0 x = 40 or x = 30 x = 30   [ Altitude cannot be negative]Thus, the altitude and base of the triangle are 30 cm and (30 + 10 = 40) cm, respectively. Hypotenuse2 = Altitude2 + Base2 Hypotenuse2 = 302 + 402 Hypotenuse2 = 900 + 1600 = 2500 Hypotenuse2 = 502 Hypotenuse = 50Thus, the dimensions of the triangle are: ,Hypotenuse = 50 cmAltitude = 30 cmBase = 40 cm

Page No 229:

Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle = 12 × Base × Altitude

 12 × 3x × x = 96  ( Area = 96 sq m) x22 = 32 x2 = 64 x = ±8


The value of cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 × 8 = 24 m), respectively.

Page No 229:

Answer:

Let the base be x m.
Therefore, the altitude will be x + 7 m.

Area of a triangle = 12 × Base × Altitude

 12 × x × (x + 7) = 165 x2 + 7x = 330 x2 + 7x  330 = 0 x2 + (22  15)x  330 = 0 x2 + 22x  15x  330 = 0 x(x + 22)  15(x + 22) = 0 (x + 22)(x  15) = 0 x = 22 or x = 15

The value of x cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

Page No 229:

Answer:

Let one side of the right-angled triangle be x m and the other side be x + 4 m.
On applying Pythagoras theorem, we have:

202 = (x + 4)2 + x2 400 = x2 + 8x + 16 + x2 2x2 + 8x  384 = 0 x2 + 4x  192 = 0 x2 + (16  12)x  192 = 0 x2 + 16x  12x  192 = 0 x(x + 16)  12(x + 16) = 0 (x + 16)(x  12) = 0 x = 16 or x = 12

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

Page No 229:

Answer:

Let the base and altitude of the right-angled triangle be x and y cm, respectively.
Therefore, the hypotenuse will be (x + 2) cm.
 (x + 2)2 = y2 + x2           ...(i)Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. h = (2y + 1) x + 2 = 2y + 1 x = 2y  1Putting the value of x in (i), we get:(2y  1 +  2)2 = y2 + (2y  1)2 (2y + 1)2 = y2 + 4y2  4y + 1 4y2 + 4y + 1 = 5y2  4y + 1 y2 + 8y = 0 y2  8y = 0 y(y   8) = 0 y = 8 cm x = 16  1 = 15 cm h = 16 + 1 = 17 cm

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

Page No 229:

Answer:

Let the shortest side be x m.
Therefore, according to the question:
Hypotenuse = 2x - 1
Third side = x + 1 m
On applying Pythagoras theorem, we get:

(2x  1)2 = (x + 1)2 + x2 4x2  4x + 1 = x2 + 2x + 1 + x2 2x2  6x = 0 2x(x  3) = 0 x = 0 or x = 3

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = 2 × 3 - 1 = 5 m
Third side = (3 + 1) = 4 m

Page No 229:

Answer:

Let the speed of faster train be x km/h.
Then, the speed of slower train is (− 10) km/h.

Given:
A faster train takes one hour less than a slower train for a journey of 200 km.

DistanceSpeed=Time

Time taken by faster train to cover 200 km = 200x h
Time taken by slower train to cover 200 km = 200x-10 h

According to the question,

200x-10-200x=1200x-200x-10xx-10=1200x-200x+2000x2-10x=12000=x2-10xx2-10x-2000=0x2-50x+40x-2000=0xx-50+40x-50=0x-50x+40=0x=50, -40But x is the speed of the train, which is always positive.Thus, x=50
and − 10 = 40

Hence, the speed of fast train is 50 km/h and the speed of slow train is 40 km/h.

Page No 229:

Answer:

Let speed of stream be x km/h.

Given:

Speed of boat = 18 km/h
Distance covered upstream = 36 km
Distance covered downstream = 36 km
It takes 32 hours more to go 36 km upstream than to return downstream to the same spot

Now, Speed of boat upstream = 18 − x km/h
Speed of boat downstream = 18 + x km/h

DistanceSpeed=Time

According to the question,

Time to go upstream = 32 hours + Time to go downstream

Time to go upstream − Time to go downstream = 32

3618-x-3618+x=323618+x-3618-x18+x18-x=32648+36x-648+36x324-x2=3272x324-x2=3224x324-x2=1224x2=324-x2x2+48x-324=0x2+54x-6x-324=0xx+54-6x+54=0x+54x-6=0x=-54, 6But x is the speed of stream which is always positive.Thus, x=6 km/h


Hence, the speed of the stream is 6 km/h.

Page No 229:

Answer:

Let the time required to fill the tank by second pipe be x hours.
Then, the time required to fill the tank by first pipe is (− 10) hours.

Given:
Two pipes together can fill a tank in 12 hours.

According to the question,

1x+1x-10=112x-10+xxx-10=1122x-10x2-10x=11224x-120=x2-10xx2-10x-24x+120=0x2-34x+120=0x2-30x-4x+120=0xx-30-4x-30=0x-30x-4=0x=30, 4But x-10 is the time required by the first pipe to fill the tank, which is always positive.Thus, x=30 and x4


Hence, the second pipe will take 30 hours to fill the tank.

Page No 229:

Answer:

Let the smaller tap takes x hours to fill the tank.
Then, the larger one takes (x − 2) hours to fill the tank.

Tank filled in 1 hour by smaller tap = 1x
Tank filled in 1 hour by larger tap = 1x-2
Tank filled in 1 hour by both the taps = 815

According to the question,

1x+1x-2=815x-2+xxx-2=8152x-2x2-2x=815152x-2=8x2-2x30x-30=8x2-16x8x2-46x+30=04x2-23x+15=0x=--23±-232-441524x=23±529-2408x=23±2898x=23±178x=23+178 or x=23-178x=5 or x=0.75But x0.75       x-2 becomes negativeThus, x=5

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.



Page No 237:

Answer:

(d) 2x2 -  5x = (x - 1)2

A quadratic equation is the equation with degree 2. 2x2  5x = (x  1)2 2x2  5x = x2  2x + 1 2x2  5x  x2 + 2x  1 = 0 x2  3x  1 = 0, which is a quadratic equation

Page No 237:

Answer:

(b) x3 - x2 = (x - 1)3

∵ x3  x2 = (x  1)3⇒ x3  x2 = x3  3x2 + 3x  1⇒ 2x2  3x + 1 = 0, which is a quadratic equation

Page No 237:

Answer:

(c) (2x + 3)2 = 2x2 + 6

∵  (2x + 3)2 = 2x2 + 6⇒ 2x2 + 9 + 62x = 2x2 + 6⇒ 62x + 3 = 0, which is not a quadratic equation

Page No 237:

Answer:

(b) −11
It is given that x=3 is a solution of 3x2+(k1)x+9=0; therefore, we have:332 + k - 1 × 3 + 9 = 0 27 + 3(k  1) + 9 = 0 3(k  1) = 36 (k  1) = 12 k = 11

Page No 237:

Answer:

(b) −7
It is given that one root of the equation 2x2 ax + 6 = 0 is 2. × 22 + a × 2 + 6 = 0 2a + 14 = 0 a = 7

Page No 237:

Answer:

(c) 6
Sum of the roots of the equation x2  6x + 2 = 0 is α β ba = (6)1 = 6 , where α and β are the roots of the equation.



Page No 238:

Answer:

(c)  8
It is given that the product of the roots of the equation x2  3x + k = 10 is 2.The equation can be rewritten as:x2 - 3x + (k - 10) = 0Product of the roots of a quadratic equation = caca = 2 (k  10)1 = 2 k = 8

Page No 238:

Answer:

(c) 2 : 3

Given:7x2  12x + 18 = 0 α + β = 127 and αβ = 187, where α and β are the roots of the equation Ratio of the sum and product of the roots = 127 : 187= 12 : 18= 2  : 3

Page No 238:

Answer:

(d) 3
Given: 3x2  10x + 3 = 0One root of the equation is 13.Let the other root be α.We know that:Product of the roots = ca 13 × α = 33 α = 3

Page No 238:

Answer:

(d) 5
Let the roots of the equation (5x2 + 13x + k = 0) be α and 1α. Product of the roots = ca α × 1α = k5 1 = k5 k = 5

Page No 238:

Answer:

(d) -23
Given:kx2 + 2x + 3k = 0Sum of the roots = Product of the roots⇒ 2k = 3kk⇒ 3k = 2⇒ k = 23

Page No 238:

Answer:

(b) x2 - 3x - 10 = 0

It is given that the roots of the quadratic equation are 5 and 2.Then, the equation is:x2  (5  2)x + 5 × (2) = 0⇒ x2  3x  10 = 0

Page No 238:

Answer:

(a) x2 - 6x + 6 = 0

Given:Sum of roots = 6Product of roots = 6Thus, the equation is: x2  6x + 6 = 0

Page No 238:

Answer:

It is given that α and β are the roots of the equation 3x2+8x+2=0.

 α+β=-83 and αβ=23
1α+1β=α+βαβ=-8323=-4

Hence, the correct answer is option C.

Page No 238:

Answer:

(c) c  =  a
Let the roots of the equation (ax2 bx + c = 0) be α and 1α. Product of the roots = α × 1α = 1 ca = 1 c = a

Page No 238:

Answer:

(d) b24a

 It is given that the roots of the equation (ax2+bx+c=0) are equal. (b2  4ac) = 0 b2 = 4ac c = b24a

Page No 238:

Answer:

(c) 2 or −2
It is given that the roots of the equation (9x2 6kx + 4 = 0) are equal. (b2  4ac) = 0 (6k)2  4 × 9 × 4 = 0 36k2 = 144 k2 = 4 k = ±2

Page No 238:

Answer:

(a) 1 or 4

It is given that the roots of the equation (x2 + 2(k + 2)x + 9k = 0) are equal. (b2  4ac) = 0 {2(k + 2)}2  4 × 1 × 9k = 0 4(k2 + 4k + 4)  36k = 0 4k2 + 16k + 16  36k = 0 4k2  20k + 16 = 0 k2  5k + 4 = 0 k2  4k  k + 4 = 0 k(k  4)  (k  4) = 0 (k  4)(k  1) = 0 k = 4 or k = 1



Page No 239:

Answer:

(d) ±43

It is given that the roots of the equation (4x2  3kx + 1 = 0) are equal. (b2  4ac) = 0 (3k)2  4 × 4 × 1 = 0 9k2 = 16 k2 = 169 k = ±43

Page No 239:

Answer:

(a) >  0
 The roots of the equation are real and unequal when (b2  4ac) > 0.

Page No 239:

Answer:

We know that when discriminant, D>0, the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

Page No 239:

Answer:

(d) imaginary

 D = (b2  4ac)= (6)2  4 × 2 × 7= 36  56= 20 < 0Thus, the roots of the equation are imaginary.

Page No 239:

Answer:

(b) real, unequal and irrational

 D = (b2  4ac)= (6)2  4 × 2 × 3= 36  24= 12 12 is greater than 0 and it is not a perfect square; therefore, the roots of the equation are real, unequal and irrational.

Page No 239:

Answer:

(d) either k > 25 or k < -25

It is given that the roots of the equation (5x2  kx + 1 = 0) are real and distinct.(b2  4ac) > 0 (k)2  4 × 5 × 1 > 0 k2  20 > 0 k2 > 20 k > 20 or k < 20 k > 25 or k < 25

Page No 239:

Answer:

(c) -85 < k > 85

It is given that the equation (x2 + 5kx + 16 = 0) has no real roots. (b2  4ac) < 0 (5k)2  4 × 1 × 16 < 0 25k2  64 < 0 k2 < 6425 85 < k < 85

Page No 239:

Answer:

(c) −2  <   <  2

It is given that the equation x2  kx + 1 = 0 has no real roots. (b2  4ac) < 0 (k)2  4 × 1 × 1 < 0 k2 < 4 2 < k < 2

Page No 239:

Answer:

(b) k  -92

It is given that the roots of the equation (kx2  6x  2 = 0) are real. D  0 (b2  4ac)  0 (6)2  4 × k × (2)  0 36 + 8k  0 k  368 k  92

Page No 239:

Answer:

(a) 54 or 45

Let the required number be x.According to the question: x + 1x = 4120 x2 + 1x = 4120 20x2  41x + 20 = 0 20x2  25x  16x + 20 = 0 5x(4x  5)  4(4x  5) = 0 (4x  5)(5x  4) = 0 x = 54 or x = 45

Page No 239:

Answer:

(c) 16 m

Let the length and breadth of the rectangle be l and b.Perimeter of the rectangle = 82 m 2 × (l + b) = 82 l + b = 41 l = (41  b)           ...(i)Area of the rectangle = 400 m2 l × b = 400 m2 (41  b)b = 400      (using (i)) 41b  b2 = 400 b2  41b + 400 = 0 b2  25b  16b + 400 = 0 b(b  25)  16(b  25) = 0 (b  25)(b  16) = 0 b = 25 or b = 16If b = 25, we have:l = 41  25 = 16 Since, l cannot be less than b, b = 16 m



Page No 240:

Answer:

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m2                    (Given)

x+8×x=240                       (Area = Length × Breadth)
x2+8x-240=0x2+20x-12x-240=0xx+20-12x+20=0x+20x-12=0
x+20=0 or x-12=0x=-20 or x=12
x = 12              (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

Page No 240:

Answer:

The given quadratic equation is 2x2-x-6=0.

2x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0x-22x+3=0
x-2=0 or 2x+3=0x=2 or x=-32
Thus, the roots of the given equation are 2 and -32.

Hence, the correct answer is option B.

Page No 240:

Answer:

Let the required natural numbers be x and (8 − x).

It is given that the product of the two numbers is 15.
x8-x=158x-x2=15x2-8x+15=0x2-5x-3x+15=0
xx-5-3x-5=0x-5x-3=0x-5=0 or x-3=0x=5 or x=3

Hence, the required numbers are 3 and 5.

Page No 240:

Answer:

The given equation is x2+6x+9=0.

Putting x = −3 in the given equation, we get

LHS = -32+6×-3+9=9-18+9=0 = RHS

∴ x = −3 is a solution of the given equation.

Page No 240:

Answer:

The given equation is 3x2+13x+14=0.

Putting x = −2 in the given equation, we get

LHS = 3×-22+13×-2+14=12-26+14=0 = RHS

∴ x = −2 is a solution of the given equation.

Page No 240:

Answer:

It is given that x=-12 is a solution of the quadratic equation 3x2+2kx-3=0.

3×-122+2k×-12-3=034-k-3=0k=3-124=-94

Hence, the value of k is -94.

Page No 240:

Answer:

The given quadratic equation is 2x2-x-6=0.
2x2-x-6=02x2-4x+3x-6=02xx-2+3x-2=0x-22x+3=0
x-2=0 or 2x+3=0x=2 or x=-32
Hence, the roots of the given equation are 2 and -32.

Page No 240:

Answer:

The given quadratic equation is 33x2+10x+3=0.
33x2+10x+3=033x2+9x+x+3=033xx+3+1x+3=0x+333x+1=0
x+3=0 or 33x+1=0x=-3 or x=-133=-39
Hence, -3 and -39 are the solutions of the given equation.

Page No 240:

Answer:

It is given that the roots of the quadratic equation 2x2+8x+k=0 are equal.
D=082-4×2×k=064-8k=0k=8
Hence, the value of k is 8.

Page No 240:

Answer:

It is given that the quadratic equation px2-25px+15=0 has two equal roots.

D=0-25p2-4×p×15=020p2-60p=020pp-3=0
p=0 or p-3=0p=0 or p=3

For p = 0, we get 15 = 0, which is not true.

p ≠ 0

Hence, the value of p is 3.

Page No 240:

Answer:

It is given that y = 1 is a root of the equation ay2+ay+3=0.
a×12+a×1+3=0a+a+3=02a+3=0a=-32

Also, y = 1 is a root of the equation y2+y+b=0.
12+1+b=01+1+b=0b+2=0b=-2

ab=-32×-2=3

Hence, the value of ab is 3.

Page No 240:

Answer:

Let the other zero of the given polynomial be α.

Now,
Sum of the zeroes of the given polynomial = --41= 4
 α+2+3=4α=4-2-3=2-3

Hence, the other zero of the given polynomial is 2-3.

Page No 240:

Answer:

Let α and β be the roots of the equation 3x2-10x+k=0.
 α=1β          (Given)
αβ=1k3=1            Product of the roots=cak=3

Hence, the value of k is 3.

Page No 240:

Answer:

It is given that the roots of the quadratic equation px2-2px+6=0 are equal.
D=0-2p2-4×p×6=04p2-24p=04pp-6=0
p=0 or p-6=0p=0 or p=6

For p = 0, we get 6 = 0, which is not true.

p ≠ 0

Hence, the value of p is 6.

Page No 240:

Answer:

It is given that the quadratic equation x2-4kx+k=0 has equal roots.
 D=0-4k2-4×1×k=016k2-4k=04k4k-1=0
k=0 or 4k-1=0k=0 or k=14

Hence, 0 and 14 are the required values of k.

Page No 240:

Answer:

It is given that the quadratic equation 9x2-3kx+k=0 has equal roots.
 D=0-3k2-4×9×k=09k2-36k=09kk-4=0
k=0 or k-4=0k=0 or k=4

Hence, 0 and 4 are the required values of k.

Page No 240:

Answer:

x2-3+1x+3=0x2-3x-x+3=0xx-3-1x-3=0x-3x-1=0
x-3=0 or x-1=0x=3 or x=1

Hence, 1 and 3 are the roots of the given equation.

Page No 240:

Answer:

2x2+ax-a2=02x2+2ax-ax-a2=02xx+a-ax+a=0x+a2x-a=0
x+a=0 or 2x-a=0x=-a or x=a2

Hence, −a and a2 are the roots of the given equation.



Page No 241:

Answer:

3x2+55x-10=03x2+65x-5x-10=03xx+25-5x+25=0x+253x-5=0
x+25=0 or 3x-5=0x=-25 or x=53

Hence, -25 and 53 are the roots of the given equation.

Page No 241:

Answer:

3x2+10x-83=03x2+12x-2x-83=03xx+43-2x+43=0x+433x-2=0
x+43=0 or 3x-2=0x=-43 or x=23=233

Hence, -43 and 233 are the roots of the given equation.

Page No 241:

Answer:

3x2-22x-23=03x2-32x+2x-23=03xx-6+2x-6=0x-63x+2=0
x-6=0 or 3x+2=0x=6 or x=-23=-63

Hence, 6 and -63 are the roots of the given equation.

Page No 241:

Answer:

43x2+5x-23=043x2+8x-3x-23=04x3x+2-33x+2=03x+24x-3=0
3x+2=0 or 4x-3=0x=-23=-233 or x=34

Hence, -233 and 34 are the roots of the given equation.

Page No 241:

Answer:

4x2+4bx-a2-b2=04x2+4bx-a-ba+b=04x2+2a+b-a-bx-a-ba+b=04x2+2a+bx-2a-bx-a-ba+b=0
2x2x+a+b-a-b2x+a+b=02x+a+b2x-a-b=02x+a+b=0 or 2x-a-b=0x=-a+b2 or x=a-b2

Hence, -a+b2 and a-b2 are the roots of the given equation.

Page No 241:

Answer:

x2+5x-a2+a-6=0x2+5x-a+3a-2=0x2+a+3-a-2x-a+3a-2=0x2+a+3x-a-2x-a+3a-2=0
xx+a+3-a-2x+a+3=0x+a+3x-a-2=0x+a+3=0 or x-a-2=0x=-a+3 or x=a-2

Hence, -a+3 and a-2 are the roots of the given equation.

Page No 241:

Answer:

x2+6x-a2+2a-8=0x2+6x-a+4a-2=0x2+a+4-a-2x-a+4a-2=0x2+a+4x-a-2x-a+4a-2=0
xx+a+4-a-2x+a+4=0x+a+4x-a-2=0x+a+4=0 or x-a-2=0x=-a+4 or x=a-2

Hence, -a+4 and a-2 are the roots of the given equation.

Page No 241:

Answer:

x2-4ax+4a2-b2=0x2-4ax+2a+b2a-b=0x2-2a+b+2a-bx+2a+b2a-b=0x2-2a+bx-2a-bx+2a+b2a-b=0
xx-2a+b-2a-bx-2a+b=0x-2a+bx-2a-b=0x-2a+b=0 or x-2a-b=0x=2a+b or x=2a-b

Hence, (2a + b) and (2ab) are the roots of the given equation.



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