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Page No 93:

Question 1:

Expand.

(i) ( 5+ 6b)2    (ii) a2 + b32   (iii) 2p - 3q2   (iv) x - 2x2

(v) ax + by2    (vi)  7m - 42    (vii)  x +122    (viii) a - 1a2

Answer:

It is known that, (a + b)2 = a2 + 2ab + b2 and (a − b)2 = a2 − 2ab + b2.
i 5a+6b2=5a2+2×5a6b+6b2=25a2+60ab+36b2
ii a2+b32=a22+2×a2×b3+b32=a24+ab3+b29
iii 2p-3q2=2p2-2×2p×3q+3q2=4p2-12pq+9q2
iv x-2x2=x2-2×x×2x+2x2=x2-4+4x2
v ax+by2=ax2+2×ax×by+by2=a2x2+2axby+b2y2
vi 7m-42=7m2-2×7m×4+42=49m2-56m+16
vii x+122=x2+2×x×12+122=x2+x+14
viii a-1a2=a2-2×a×1a+1a2=a2-2+1a2

Page No 93:

Question 2:

Which of the options given below is the square of the binomial 8 - 1x ?

(i) 64 - 1x2    (ii) 64 +1x2  (iii)  64 - 16x + 1x2    (iv) 64 +16x + 1x2

Answer:


The given binomial is 8-1x.
8-1x2=82-2×8×1x+1x2     a-b2=a2-2ab+b2=64-16x+1x2
Hence, the correct answer is option (iii).

Page No 93:

Question 3:

Of which of the binomials given below is m2n2 + 14mnpq + 49p2q2 the expansion?
 
(i) ( + n) (p + q) (ii) (mn- pq) (iii) (7mn + pq) (iv) (mn + 7pq)

Answer:


Let us check each of the given options.
(i) (m + n)(p + q)
= m(p + q) + n(p + q)
mp + mq + np + nq
So, it is not a correct option.

(ii) (mn − pq)2
= (mn)2 − 2 × (mn) × (pq) + (pq)2    [∵ (a − b)2 = a2 − 2ab + b2]
= m2n2 − 2mnpq + p2q2
So, it is not a correct option.

(iii) (7mn + pq)2
= (7mn)2 + 2 × (7mn) × (pq) + (pq)2    [∵ (a + b)2 = a2 + 2ab + b2]
= 49m2n2 + 14mnpq + p2q2
So, it is not a correct option.

(iv) (mn + 7pq)2
= (mn)2 + 2 × (mn) × (7pq) + (7pq)2    [∵ (a + b)2 = a2 + 2ab + b2]
= m2n2 + 14mnpq + 49p2q2
So, it is a correct option.

Hence, the correct answer is option (iv).

Page No 93:

Question 4:

Use an expansion formula to find the values.
(i) (997)2 (ii) (102)2 (iii) (97)2 (iv) (1005)2

Answer:

It is known that, (a + b)2 = a2 + 2ab + b2 and (a − b)2 = a2 − 2ab + b2

(i) (997)2
= (1000 − 3)2
= (1000)2 − 2 × 1000 × 3 + (3)2
= 1000000 − 6000 + 9
= 994009

(ii) (102)2
= (100 + 2)2
= (100)2 + 2 × 100 × 2 + (2)2
= 10000 + 400 + 4
= 10404

(iii) (97)2
= (100 − 3)2
= (100)2 − 2 × 100 × 3 + (3)2
= 10000 − 600 + 9
= 9409

(iv) (1005)2
= (1000 + 5)2
= (1000)2 + 2 × 1000 × 5 + (5)2
= 1000000 + 10000 + 25
= 1010025

Page No 93:

Question 1:

Use the formula to multiply the following.
(i) (x + y) (x -y) (ii) (3x-5) (3x + 5)  (iii) (a + 6) (a -6) (iv) x5 + 6 x5 - 6

Answer:

It is known that, (a + b) (a − b) = a2 − b2.
i x+yx-y=x2-y2=x2-y2
ii 3x-53x+5=3x2-52=9x2-25
iii a+6a-6=a2-62=a2-36
iv x5+6x5-6=x52-62=x225-36

Page No 93:

Question 2:

Use the formula to find the values.
(i) 502 × 498 (ii) 97 × 103 (iii) 54 × 46 (iv) 98 × 102

 

Answer:

It is known that, (a + b) (a − b) = a2 − b2.

(i) 502 × 498
= (500 + 2) × (500 − 2)
= (500)2 −  (2)2
= 250000 − 4
= 249996

(ii) 97 × 103
= (100 − 3) × (100 + 3)
= (100)2 − (3)2
= 10000 − 9
= 9991

(iii) 54 × 46
= (50 + 4) × (50 − 4)
= (50)2 − (4)2
= 2500 − 16
= 2484

(iv) 98 × 102
= (100 − 2) × (100 + 2)
= (100)2 − (2)2
= 10000 − 4
= 9996



Page No 94:

Question 1:

Factorise the following expressions and write them in the product form.
 
(i) 201 a3 b2 , (ii) 91 xyt2 , (iii) 24 a2 b2 , (iv) tr2 s3

Answer:


(i) 201a3b2
= 3 × 67 × a × a × a × b × b

(ii) 91xyt2
= 7 × 13 × x × y × t × t

(iii) 24a2b2
= 2 × 2 × 2 × 3 × a × a × b × b

(iv) tr2s3
= t × × r × s × s × s

Page No 94:

Question 1:

Factorise the following expressions.
(i) p 2-q 2         (ii) 4x2 - 25y2  (iii) y2 - 4        (iv) p- 125
(v) 9x-116y2 (vi) x2 - 1x2     (vii) a2b -ab   (viii) 4 x2y -x2
(ix) 12y2 - 8z2  (x) 2x2 - 8y2
 

Answer:


i p2-q2=p2-q2=p+qp-q     a+ba-b=a2-b2
(ii) 4x2-25y2=2x2-5y2=2x+5y2x-5y     a2-b2=a+ba-b
iii y2-4=y2-22=y+2y-2     a2-b2=a+ba-b
iv p2-125=p2-152=p+15p-15     a2-b2=a+ba-b
v 9x2-116y2=3x2-14y2=3x+14y3x-14y     a2-b2=a+ba-b
vi x2-1x2=x2-1x2=x+1xx-1x     a2-b2=a+ba-b
vii a2b-ab=aba-1
viii 4x2y-6x2=2x22y-3
ix 12y2-8z2=12y2-16z2=12y2-4z2=12y+4zy-4z     a2-b2=a+ba-b
x 2x2-8y2=2x2-4y2=2x2-2y2=2x+2yx-2y
 



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