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 Meritnationâ€™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:

(ix) ${\left(x+2\right)}^{3}={x}^{3}-8$
$⇒{x}^{3}+6{x}^{2}+12x+8={x}^{3}-8\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+12x+16=0$
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $\left(2x+3\right)\left(3x+2\right)=6\left(x-1\right)\left(x-2\right)$
$⇒6{x}^{2}+4x+9x+6=6\left({x}^{2}-3x+2\right)\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+13x+6=6{x}^{2}-18x+12\phantom{\rule{0ex}{0ex}}⇒31x-6=0$
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${\left(x+\frac{1}{x}\right)}^{2}=2\left(x+\frac{1}{x}\right)+3$
$⇒{\left(\frac{{x}^{2}+1}{x}\right)}^{2}=2\left(\frac{{x}^{2}+1}{x}\right)+3\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}+1\right)}^{2}=2x\left({x}^{2}+1\right)+3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{4}+2{x}^{2}+1=2{x}^{3}+2x+3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{4}-2{x}^{3}-{x}^{2}-2x+1=0$
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

#### Page No 182:

(i)

So, the equation becomes ${x}^{2}-4x+3=0$
On factorising we get;

Hence, the other root is 3.

(ii)

#### Page No 182:

LHS;
Consider the quadratic equation;
$a{d}^{2}\left(\frac{ax}{b}+\frac{2c}{d}\right)x+b{c}^{2}=0$
Put $x=-\frac{bc}{ad}$ in the given equation.
$a{d}^{2}\left[\frac{a\left(-\frac{bc}{ad}\right)}{b}+\frac{2c}{d}\right]\left(-\frac{bc}{ad}\right)+b{c}^{2}\phantom{\rule{0ex}{0ex}}=\frac{a{d}^{2}}{bd}\left[ad\left(-\frac{bc}{ad}\right)+2bc\right]\left(-\frac{bc}{ad}\right)+b{c}^{2}\phantom{\rule{0ex}{0ex}}=\frac{a{d}^{2}}{bd}\left[-bc+2bc\right]\left(-\frac{bc}{ad}\right)+b{c}^{2}\phantom{\rule{0ex}{0ex}}=\frac{a{d}^{2}}{bd}\left(bc\right)\left(-\frac{bc}{ad}\right)+b{c}^{2}\phantom{\rule{0ex}{0ex}}=-\frac{a{b}^{2}{c}^{2}{d}^{2}}{ab{d}^{2}}+b{c}^{2}\phantom{\rule{0ex}{0ex}}=-b{c}^{2}+b{c}^{2}\phantom{\rule{0ex}{0ex}}=0=\mathrm{RHS}$
Hence, $x=-\frac{bc}{ad}$ is a solution to the given quadratic equation.

#### Page No 182:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

x$\frac{3}{2}$ or x = $-\frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $-\frac{1}{3}$.

#### Page No 182:

4x2 + 5x = 0

x(4x + 5) = 0

x = 0 or 4x + 5 = 0

x = 0 or x = $-\frac{5}{4}$

Hence, the roots of the given equation are 0 and $-\frac{5}{4}$.

#### Page No 182:

We write, $x=4x-3x$ as $2{x}^{2}×\left(-6\right)=-12{x}^{2}=4x×\left(-3x\right)$

Hence, the roots of the given equation are $-2$ and $\frac{3}{2}$.

#### Page No 182:

We write, $6x=x+5x$ as ${x}^{2}×5=5{x}^{2}=x×5x$

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

#### Page No 182:

We write, $-3x=3x-6x$ as $9{x}^{2}×\left(-2\right)=-18{x}^{2}=3x×\left(-6x\right)$

Hence, the roots of the given equation are $-\frac{1}{3}$ and $\frac{2}{3}$.

#### Page No 183:

We write, $-2x=-3x+x$ as $3{x}^{2}×\left(-1\right)=-3{x}^{2}=\left(-3x\right)×x$

Hence, the roots of the given equation are $1$ and $-\frac{1}{3}$.

#### Page No 183:

We write, $2\sqrt{2}x=3\sqrt{2}x-\sqrt{2}x$ as ${x}^{2}×\left(-6\right)=-6{x}^{2}=3\sqrt{2}x×\left(-\sqrt{2}x\right)$

Hence, the roots of the given equation are $-3\sqrt{2}$ and $\sqrt{2}$.

#### Page No 183:

Consider $\sqrt{3}{x}^{2}+10x-8\sqrt{3}=0$
Factorising by splitting the middle term;

Hence, the roots of the given equation are  and $-4\sqrt{3}$.

#### Page No 183:

We write, $-6x=7x-13x$ as $\sqrt{7}{x}^{2}×\left(-13\sqrt{7}\right)=-91{x}^{2}=7x×\left(-13x\right)$

$\therefore \sqrt{7}{x}^{2}-6x-13\sqrt{7}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}{x}^{2}+7x-13x-13\sqrt{7}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}x\left(x+\sqrt{7}\right)-13\left(x+\sqrt{7}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+\sqrt{7}\right)\left(\sqrt{7}x-13\right)=0$

Hence, the roots of the given equation are $-\sqrt{7}$ and $\frac{13\sqrt{7}}{7}$.

#### Page No 183:

We write, $-2\sqrt{6}x=-\sqrt{6}x-\sqrt{6}x$ as $3{x}^{2}×2=6{x}^{2}=\left(-\sqrt{6}x\right)×\left(-\sqrt{6}x\right)$

$\therefore 3{x}^{2}-2\sqrt{6}x+2=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-\sqrt{6}x-\sqrt{6}x+2=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}x-\sqrt{2}\right)\left(\sqrt{3}x-\sqrt{2}\right)=0$

Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

#### Page No 183:

We write, $-2\sqrt{2}x=-3\sqrt{2}x+\sqrt{2}x$ as $\sqrt{3}{x}^{2}×\left(-2\sqrt{3}\right)=-6{x}^{2}=\left(-3\sqrt{2}x\right)×\left(\sqrt{2}x\right)$

$\therefore \sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}{x}^{2}-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(x-\sqrt{6}\right)+\sqrt{2}\left(x-\sqrt{6}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-\sqrt{6}\right)\left(\sqrt{3}x+\sqrt{2}\right)=0$

Hence, the roots of the given equation are $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$.

#### Page No 183:

We write, $-3\sqrt{5}x=-2\sqrt{5}x-\sqrt{5}x$ as ${x}^{2}×10=10{x}^{2}=\left(-2\sqrt{5}x\right)×\left(-\sqrt{5}x\right)$

$\therefore {x}^{2}-3\sqrt{5}x+10=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2\sqrt{5}x-\sqrt{5}x+10=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-2\sqrt{5}\right)-\sqrt{5}\left(x-2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\sqrt{5}\right)\left(x-\sqrt{5}\right)=0$

Hence, the roots of the given equation are $\sqrt{5}$ and $2\sqrt{5}$.

#### Page No 183:

${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\sqrt{3}x-x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-\sqrt{3}\right)-1\left(x-\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-\sqrt{3}\right)\left(x-1\right)=0$

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

#### Page No 183:

We write, $3\sqrt{3}x=5\sqrt{3}x-2\sqrt{3}x$ as ${x}^{2}×\left(-30\right)=-30{x}^{2}=5\sqrt{3}x×\left(-2\sqrt{3}x\right)$

$\therefore {x}^{2}+3\sqrt{3}x-30=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5\sqrt{3}x-2\sqrt{3}x-30=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+5\sqrt{3}\right)-2\sqrt{3}\left(x+5\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+5\sqrt{3}\right)\left(x-2\sqrt{3}\right)=0$

Hence, the roots of the given equation are $-5\sqrt{3}$ and $2\sqrt{3}$.

#### Page No 183:

We write, $7x=5x+2x$ as $\sqrt{2}{x}^{2}×5\sqrt{2}=10{x}^{2}=5x×2x$

$\therefore \sqrt{2}{x}^{2}+7x+5\sqrt{2}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{2}{x}^{2}+5x+2x+5\sqrt{2}=0\phantom{\rule{0ex}{0ex}}⇒x\left(\sqrt{2}x+5\right)+\sqrt{2}\left(\sqrt{2}x+5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{2}x+5\right)\left(x+\sqrt{2}\right)=0$

Hence, the roots of the given equation are $-\sqrt{2}$ and $-\frac{5\sqrt{2}}{2}$.

#### Page No 183:

We write, 13x = 5x + 8x as $5{x}^{2}×8=40{x}^{2}=5x×8x$

Hence, $-1$ and $-\frac{8}{5}$ are the roots of the given equation.

#### Page No 183:

We write, $-20x=-10x-10x$ as $100{x}^{2}×1=100{x}^{2}=\left(-10x\right)×\left(-10x\right)$

$⇒{\left(10x-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒10x-1=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

#### Page No 183:

We write, $-x=-\frac{x}{2}-\frac{x}{2}$ as $2{x}^{2}×\frac{1}{8}=\frac{{x}^{2}}{4}=\left(-\frac{x}{2}\right)×\left(-\frac{x}{2}\right)$

Hence, $\frac{1}{4}$ is the repeated root of the given equation.

#### Page No 183:

We write, $ax=2ax-ax$ as $2{x}^{2}×\left(-{a}^{2}\right)=-2{a}^{2}{x}^{2}=2ax×\left(-ax\right)$

Hence, $-a$ and $\frac{a}{2}$ are the roots of the given equation.

#### Page No 183:

We write, $4bx=2\left(a+b\right)x-2\left(a-b\right)x$ as $4{x}^{2}×\left[-\left({a}^{2}-{b}^{2}\right)\right]=-4\left({a}^{2}-{b}^{2}\right){x}^{2}=2\left(a+b\right)x×\left[-2\left(a-b\right)x\right]$

Hence, $-\frac{a+b}{2}$ and $\frac{a-b}{2}$ are the roots of the given equation.

#### Page No 183:

We write, $-4{a}^{2}x=-2\left({a}^{2}+{b}^{2}\right)x-2\left({a}^{2}-{b}^{2}\right)x$ as $4{x}^{2}×\left({a}^{4}-{b}^{4}\right)=4\left({a}^{4}-{b}^{4}\right){x}^{2}=\left[-2\left({a}^{2}+{b}^{2}\right)\right]x×\left[-2\left({a}^{2}-{b}^{2}\right)\right]x$

Hence, $\frac{{a}^{2}+{b}^{2}}{2}$ and $\frac{{a}^{2}-{b}^{2}}{2}$ are the roots of the given equation.

#### Page No 183:

We write, $5x=\left(a+3\right)x-\left(a-2\right)x$ as ${x}^{2}×\left[-\left({a}^{2}+a-6\right)\right]=-\left({a}^{2}+a-6\right){x}^{2}=\left(a+3\right)x×\left[-\left(a-2\right)x\right]$

Hence, $-\left(a+3\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Page No 183:

We write, $-2ax=\left(2b-a\right)x-\left(2b+a\right)x$ as ${x}^{2}×\left[-\left(4{b}^{2}-{a}^{2}\right)\right]=-\left(4{b}^{2}-{a}^{2}\right){x}^{2}=\left(2b-a\right)x×\left[-\left(2b+a\right)x\right]$

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

#### Page No 183:

We write, $-\left(2b-1\right)x=-\left(b-5\right)x-\left(b+4\right)x$ as ${x}^{2}×\left({b}^{2}-b-20\right)=\left({b}^{2}-b-20\right){x}^{2}=\left[-\left(b-5\right)x\right]×\left[-\left(b+4\right)x\right]$

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

#### Page No 183:

We write, $6x=\left(a+4\right)x-\left(a-2\right)x$ as ${x}^{2}×\left[-\left({a}^{2}+2a-8\right)\right]=-\left({a}^{2}+2a-8\right){x}^{2}=\left(a+4\right)x×\left[-\left(a-2\right)x\right]$

Hence, $-\left(a+4\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Page No 183:

We write, $-4ax=-\left(b+2a\right)x+\left(b-2a\right)x$ as ${x}^{2}×\left(-{b}^{2}+4{a}^{2}\right)=\left(-{b}^{2}+4{a}^{2}\right){x}^{2}=-\left(b+2a\right)x×\left(b-2a\right)x$

Hence, $\left(2a+b\right)$ and $\left(2a-b\right)$ are the roots of the given equation.

#### Page No 183:

We write, $-9\left(a+b\right)x=-3\left(2a+b\right)x-3\left(a+2b\right)x$ as $9{x}^{2}×\left(2{a}^{2}+5ab+2{b}^{2}\right)=9\left(2{a}^{2}+5ab+2{b}^{2}\right){x}^{2}=\left[-3\left(2a+b\right)x\right]×\left[-3\left(a+2b\right)x\right]$
$\therefore 9{x}^{2}-9\left(a+b\right)x+\left(2{a}^{2}+5ab+2{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-3\left(2a+b\right)x-3\left(a+2b\right)x+\left(2a+b\right)\left(a+2b\right)=0\phantom{\rule{0ex}{0ex}}⇒3x\left[3x-\left(2a+b\right)\right]-\left(a+2b\right)\left[3x-\left(2a+b\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒\left[3x-\left(2a+b\right)\right]\left[3x-\left(a+2b\right)\right]=0$

Hence, $\frac{2a+b}{3}$ and $\frac{a+2b}{3}$ are the roots of the given equation.

#### Page No 183:

Hence, −4 and 4 are the roots of the given equation.

#### Page No 183:

Hence, −2 and 1 are the roots of the given equation.

#### Page No 183:

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

#### Page No 184:

(i)

$⇒{x}^{2}+4x-12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+6x-2x-12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+6\right)-2\left(x+6\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+6\right)\left(x-2\right)=0$

Hence, −6 and 2 are the roots of the given equation.

(ii)

#### Page No 184:

$\frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2a+b+2x}-\frac{1}{2x}=\frac{1}{2a}+\frac{1}{b}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-2a-b-2x}{2x\left(2a+b+2x\right)}=\frac{2a+b}{2ab}\phantom{\rule{0ex}{0ex}}⇒\frac{-\left(2a+b\right)}{4{x}^{2}+4ax+2bx}=\frac{2a+b}{2ab}$
$⇒4{x}^{2}+4ax+2bx=-2ab\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+4ax+2bx+2ab=0\phantom{\rule{0ex}{0ex}}⇒4x\left(x+a\right)+2b\left(x+a\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+a\right)\left(4x+2b\right)=0$

Hence, $-a$ and $-\frac{b}{2}$ are the roots of the given equation.

#### Page No 184:

$⇒18{x}^{2}-48x+130=105x-140\phantom{\rule{0ex}{0ex}}⇒18{x}^{2}-153x+270=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-17x+30=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-12x-5x+30=0$

Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

#### Page No 184:

(i)

$⇒8{x}^{2}-8x+4=17{x}^{2}-17x\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-9x-4=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-12x+3x-4=0\phantom{\rule{0ex}{0ex}}⇒3x\left(3x-4\right)+1\left(3x-4\right)=0$

Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.

(ii)

#### Page No 184:

$⇒30{x}^{2}+30x+15=34{x}^{2}+34x\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+4x-15=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+10x-6x-15=0\phantom{\rule{0ex}{0ex}}⇒2x\left(2x+5\right)-3\left(2x+5\right)=0$

Hence, $-\frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

#### Page No 184:

$⇒\frac{{x}^{2}-11x+29}{{x}^{2}-12x+35}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-33x+87=5{x}^{2}-60x+175\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-27x+88=0$
$⇒2{x}^{2}-16x-11x+88=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-8\right)-11\left(x-8\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-8\right)\left(2x-11\right)=0$

Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

#### Page No 184:

$⇒\frac{{x}^{2}-5x+5}{{x}^{2}-6x+8}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-15x+15=5{x}^{2}-30x+40\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-15x+25=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-10x-5x+25=0$

Hence, 5 and $\frac{5}{2}$ are the roots of the given equation.

#### Page No 184:

(i)

$⇒3{x}^{2}+16x+16=5{x}^{2}+15x+10\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0$

Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.

(ii)

#### Page No 184:

$⇒\frac{19{x}^{2}-42x-15}{6{x}^{2}+7x-3}=5\phantom{\rule{0ex}{0ex}}⇒19{x}^{2}-42x-15=30{x}^{2}+35x-15\phantom{\rule{0ex}{0ex}}⇒11{x}^{2}+77x=0\phantom{\rule{0ex}{0ex}}⇒11x\left(x+7\right)=0$

Hence, 0 and −7 are the roots of the given equation.

#### Page No 184:

$⇒\frac{47{x}^{2}+162x-33}{35{x}^{2}-16x-3}=11\phantom{\rule{0ex}{0ex}}⇒47{x}^{2}+162x-33=385{x}^{2}-176x-33\phantom{\rule{0ex}{0ex}}⇒338{x}^{2}-338x=0\phantom{\rule{0ex}{0ex}}⇒338x\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}$

Hence, 0 and 1 are the roots of the given equation.

#### Page No 193:

â€‹

â€‹

(v) $\left(x-1\right)\left(2x-1\right)=0$

$⇒2{x}^{2}-3x+1=0$

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×2×1=9-8=1$

â€‹

#### Page No 193:

The given equation is $2{x}^{2}+x-4=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = ${b}^{2}-4ac={\left(1\right)}^{2}-4×2×\left(-4\right)=1+32=33>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{33}$

Hence, $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$ are the roots of the given equation.

#### Page No 193:

The given equation is $2{x}^{2}-2\sqrt{2}x+1=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$-2\sqrt{2}$ and c = 1

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{2}\right)}^{2}-4×2×1=8-8=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

#### Page No 193:

The given equation is $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $\sqrt{2}$, b = 7 and c = $5\sqrt{2}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(7\right)}^{2}-4×\sqrt{2}×5\sqrt{2}=49-40=9>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{9}=3$

Hence, $-\sqrt{2}$ and $-\frac{5\sqrt{2}}{2}$ are the roots of the given equation.

#### Page No 193:

The given equation is $\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $\sqrt{3}$, b$-2\sqrt{2}$ and c = $-2\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{2}\right)}^{2}-4×\sqrt{3}×\left(-2\sqrt{3}\right)=8+24=32>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{32}=4\sqrt{2}$

Hence, $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is $2{x}^{2}+6\sqrt{3}x-60=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$6\sqrt{3}$ and c = $-60$

∴ Discriminant, D = ${b}^{2}-4ac={\left(6\sqrt{3}\right)}^{2}-4×2×\left(-60\right)=108+480=588>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{588}=14\sqrt{3}$

Hence, $2\sqrt{3}$ and $-5\sqrt{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is $4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $4\sqrt{3}$, b = 5 and c = $-2\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={5}^{2}-4×4\sqrt{3}×\left(-2\sqrt{3}\right)=25+96=121>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{121}=11$

Hence, $\frac{\sqrt{3}}{4}$ and $-\frac{2\sqrt{3}}{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is $3{x}^{2}-2\sqrt{6}x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 3, b$-2\sqrt{6}$ and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{6}\right)}^{2}-4×3×2=24-24=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

#### Page No 193:

The given equation is $2\sqrt{3}{x}^{2}-5x+\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $2\sqrt{3}$, b$-5$ and c = $\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(-5\right)}^{2}-4×2\sqrt{3}×\sqrt{3}=25-24=1>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{1}=1$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is ${x}^{2}+x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={1}^{2}-4×1×2=1-8=-7<0$

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

#### Page No 193:

The given equation is $2{x}^{2}+ax-{a}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 2, B = a and C = $-{a}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={a}^{2}-4×2×-{a}^{2}={a}^{2}+8{a}^{2}=9{a}^{2}\ge 0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{9{a}^{2}}=3a$

Hence, $\frac{a}{2}$ and $-a$ are the roots of the given equation.

#### Page No 193:

The given equation is ${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 1, b$-\left(\sqrt{3}+1\right)$ and c = $\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left[-\left(\sqrt{3}+1\right)\right]}^{2}-4×1×\sqrt{3}=3+1+2\sqrt{3}-4\sqrt{3}=3-2\sqrt{3}+1={\left(\sqrt{3}-1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(\sqrt{3}-1\right)}^{2}}=\sqrt{3}-1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

#### Page No 193:

The given equation is $2{x}^{2}+5\sqrt{3}x+6=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$5\sqrt{3}$ and c = 6

∴ Discriminant, D = ${b}^{2}-4ac={\left(5\sqrt{3}\right)}^{2}-4×2×6=75-48=27>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{27}=3\sqrt{3}$

Hence, $-\frac{\sqrt{3}}{2}$ and $-2\sqrt{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is $3{x}^{2}-2x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\right)}^{2}-4×3×2=4-24=-20<0$

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

#### Page No 193:

The given equation is

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −3 and c = 1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×1×1=9-4=5>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{5}$

Hence, $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the roots of the given equation.

#### Page No 193:

The given equation is

$⇒3{x}^{2}-6x+2=0$

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 3, b = −6 and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-6\right)}^{2}-4×3×2=36-24=12>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{12}=2\sqrt{3}$

Hence, $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$ are the roots of the given equation.

#### Page No 193:

The given equation is

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −3 and c = −1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×1×\left(-1\right)=9+4=13>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{13}$

Hence, $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$ are the roots of the given equation.

#### Page No 193:

The given equation is

$\frac{m}{n}{x}^{2}+\frac{n}{m}=1-2x\phantom{\rule{0ex}{0ex}}⇒\frac{{m}^{2}{x}^{2}+{n}^{2}}{mn}=1-2x\phantom{\rule{0ex}{0ex}}⇒{m}^{2}{x}^{2}+{n}^{2}=mn-2mnx\phantom{\rule{0ex}{0ex}}⇒{m}^{2}{x}^{2}+2mnx+{n}^{2}-mn=0$

This equation is of the form $a{x}^{2}+bx+c=0$, where a = ${m}^{2}$, b = 2mn and c = ${n}^{2}-mn$.

∴ Discriminant, D = ${b}^{2}-4ac={\left(2mn\right)}^{2}-4×{m}^{2}×\left({n}^{2}-mn\right)=4{m}^{2}{n}^{2}-4{m}^{2}{n}^{2}+4{m}^{3}n=4{m}^{3}n>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{m}^{3}n}=2m\sqrt{mn}$

Hence, $\frac{-n+\sqrt{mn}}{m}$ and $\frac{-n-\sqrt{mn}}{m}$ are the roots of the given equation.

#### Page No 193:

The given equation is $36{x}^{2}-12ax+\left({a}^{2}-{b}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 36, B$-12a$ and C = ${a}^{2}-{b}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-12a\right)}^{2}-4×36×\left({a}^{2}-{b}^{2}\right)=144{a}^{2}-144{a}^{2}+144{b}^{2}=144{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{144{b}^{2}}=12b$

Hence, $\frac{a+b}{6}$ and $\frac{a-b}{6}$ are the roots of the given equation.

#### Page No 193:

The given equation is ${x}^{2}-2ax-\left(4{b}^{2}-{a}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B$-2a$ and C = $-\left(4{b}^{2}-{a}^{2}\right)$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-2a\right)}^{2}-4×1×\left[-\left(4{b}^{2}-{a}^{2}\right)\right]=4{a}^{2}+16{b}^{2}-4{a}^{2}=16{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{b}^{2}}=4b$

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

#### Page No 194:

The given equation is ${x}^{2}+6x-\left({a}^{2}+2a-8\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = 6 and C = $-\left({a}^{2}+2a-8\right)$

∴ Discriminant, D = ${B}^{2}-4AC={6}^{2}-4×1×\left[-\left({a}^{2}+2a-8\right)\right]=36+4{a}^{2}+8a-32=4{a}^{2}+8a+4=4\left({a}^{2}+2a+1\right)=4{\left(a+1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{\left(a+1\right)}^{2}}=2\left(a+1\right)$

Hence, $\left(a-2\right)$ and $-\left(a+4\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is ${x}^{2}+5x-\left({a}^{2}+a-6\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = 5 and C = $-\left({a}^{2}+a-6\right)$

∴ Discriminant, D = ${B}^{2}-4AC={5}^{2}-4×1×\left[-\left({a}^{2}+a-6\right)\right]=25+4{a}^{2}+4a-24=4{a}^{2}+4a+1={\left(2a+1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(2a+1\right)}^{2}}=2a+1$

Hence, $\left(a-2\right)$ and $-\left(a+3\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is ${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = −4a and C = $-{b}^{2}+4{a}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-4a\right)}^{2}-4×1×\left(-{b}^{2}+4{a}^{2}\right)=16{a}^{2}+4{b}^{2}-16{a}^{2}=4{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{b}^{2}}=2b$

Hence, $\left(2a+b\right)$ and $\left(2a-b\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is $4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 4, B = −4a2 and C = ${a}^{4}-{b}^{4}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-4{a}^{2}\right)}^{2}-4×4×\left({a}^{4}-{b}^{4}\right)=16{a}^{4}-16{a}^{4}+16{b}^{4}=16{b}^{4}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{b}^{4}}=4{b}^{2}$

Hence, $\frac{1}{2}\left({a}^{2}+{b}^{2}\right)$ and $\frac{1}{2}\left({a}^{2}-{b}^{2}\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is $4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 4, B = 4b and C = $-\left({a}^{2}-{b}^{2}\right)$

∴ Discriminant, D = ${B}^{2}-4AC={\left(4b\right)}^{2}-4×4×\left[-\left({a}^{2}-{b}^{2}\right)\right]=16{b}^{2}+16{a}^{2}-16{b}^{2}=16{a}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{a}^{2}}=4a$

Hence, $\frac{1}{2}\left(a-b\right)$ and $-\frac{1}{2}\left(a+b\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is ${x}^{2}-\left(2b-1\right)x+\left({b}^{2}-b-20\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = $-\left(2b-1\right)$ and C = ${b}^{2}-b-20$

∴ Discriminant, D = ${B}^{2}-4AC={\left[-\left(2b-1\right)\right]}^{2}-4×1×\left({b}^{2}-b-20\right)=4{b}^{2}-4b+1-4{b}^{2}+4b+80=81>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{81}=9$

Hence, $\left(b+4\right)$ and $\left(b-5\right)$ are the roots of the given equation.

#### Page No 194:

The given equation is ${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = ${a}^{2}{b}^{2}$, B = $-\left(4{b}^{4}-3{a}^{4}\right)$ and C = $-12{a}^{2}{b}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left[-\left(4{b}^{4}-3{a}^{4}\right)\right]}^{2}-4×{a}^{2}{b}^{2}×\left(-12{a}^{2}{b}^{2}\right)=16{b}^{8}-24{a}^{4}{b}^{4}+9{a}^{8}+48{a}^{4}{b}^{4}=16{b}^{8}+24{a}^{4}{b}^{4}+9{a}^{8}={\left(4{b}^{4}+3{a}^{4}\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(4{b}^{4}+3{a}^{4}\right)}^{2}}=4{b}^{4}+3{a}^{4}$

Hence, $\frac{4{b}^{2}}{{a}^{2}}$ and $-\frac{3{a}^{2}}{{b}^{2}}$ are the roots of the given equation.

#### Page No 201:

(i) The given equation is $2{x}^{2}-8x+5=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 2, b = −8 and c = 5.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-8\right)}^{2}-4×2×5=64-40=24>0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3{x}^{2}-2\sqrt{6}x+2=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 3, b$-2\sqrt{6}$ and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{6}\right)}^{2}-4×3×2=24-24=0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5{x}^{2}-4x+1=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 5, b = −4 and c = 1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-4\right)}^{2}-4×5×1=16-20=-4<0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5x\left(x-2\right)+6=0\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}-10x+6=0$

This is of the form $a{x}^{2}+bx+c=0$, where a = 5, b = −10 and c = 6.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-10\right)}^{2}-4×5×6=100-120=-20<0$

Hence, the given equation has no real roots.

(v) The given equation is $12{x}^{2}-4\sqrt{15}x+5=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 12, b$-4\sqrt{15}$ and c = 5.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-4\sqrt{15}\right)}^{2}-4×12×5=240-240=0$

Hence, the given equation has real and equal roots.

(vi) The given equation is ${x}^{2}-x+2=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −1 and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-1\right)}^{2}-4×1×2=1-8=-7<0$

Hence, the given equation has no real roots.

#### Page No 201:

The given equation is $2\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(a+b\right)x+1=0$.

Hence, the given equation has no real roots.

#### Page No 202:

The given equation is

$kx\left(x-2\sqrt{5}\right)+10=0\phantom{\rule{0ex}{0ex}}⇒k{x}^{2}-2\sqrt{5}kx+10=0$

This is of the form $a{x}^{2}+bx+c=0$, where a = k, b = $-2\sqrt{5}k$ and c = 10.

$\therefore D={b}^{2}-4ac={\left(-2\sqrt{5}k\right)}^{2}-4×k×10=20{k}^{2}-40k$

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

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

Hence, 2 is the required value of k.

#### Page No 202:

The given equation is $4{x}^{2}+px+3=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 4, b = p and c = 3.

$\therefore D={b}^{2}-4ac={p}^{2}-4×4×3={p}^{2}-48$

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

$\therefore {p}^{2}-48=0\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=48\phantom{\rule{0ex}{0ex}}⇒p=±\sqrt{48}=±4\sqrt{3}$

Hence, $4\sqrt{3}$ and $-4\sqrt{3}$ are the required values of p.

#### Page No 202:

The given equation is $9{x}^{2}-3kx+k=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 9, b = −3k and c = k.

$\therefore D={b}^{2}-4ac={\left(-3k\right)}^{2}-4×9×k=9{k}^{2}-36k$

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

But, k ≠ 0        (Given)

Hence, the required value of k is 4.

#### Page No 202:

(i)
The given equation is $\left(3k+1\right){x}^{2}+2\left(k+1\right)x+1=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 3k +1, b = 2(k + 1) and c = 1.

$=4{k}^{2}-4k$

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

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

(ii)

#### Page No 202:

The given equation is $\left(2p+1\right){x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$.

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

$=-7{p}^{2}+24p+16$

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

$\therefore -7{p}^{2}+24p+16=0\phantom{\rule{0ex}{0ex}}⇒7{p}^{2}-24p-16=0\phantom{\rule{0ex}{0ex}}⇒7{p}^{2}-28p+4p-16=0\phantom{\rule{0ex}{0ex}}⇒7p\left(p-4\right)+4\left(p-4\right)=0$

Hence, 4 and $-\frac{4}{7}$ are the required values of p.

#### Page No 202:

The given equation is $\left(p+1\right){x}^{2}-6\left(p+1\right)x+3\left(p+9\right)=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = p +1, b = 6(p + 1) and c = 3(p + 9).

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

But, $p\ne -1$           (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4{x}^{2}-24x+36=0$.

$4{x}^{2}-24x+36=0\phantom{\rule{0ex}{0ex}}⇒4\left({x}^{2}-6x+9\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-3\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒x-3=0$
$⇒x=3$

Hence, 3 is the repeated root of this equation.

#### Page No 202:

It is given that −5 is a root of the quadratic equation $2{x}^{2}+px-15=0$.

$\therefore 2{\left(-5\right)}^{2}+p×\left(-5\right)-15=0\phantom{\rule{0ex}{0ex}}⇒-5p+35=0\phantom{\rule{0ex}{0ex}}⇒p=7$

The roots of the equation $p{x}^{2}+px+k=0$ = 0 are equal.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{p}^{2}-4pk=0\phantom{\rule{0ex}{0ex}}⇒{\left(7\right)}^{2}-4×7×k=0\phantom{\rule{0ex}{0ex}}⇒49-28k=0$
$⇒k=\frac{49}{28}=\frac{7}{4}$

Thus, the value of k is $\frac{7}{4}$.

#### Page No 202:

It is given that 3 is a root of the quadratic equation ${x}^{2}-x+k=0$.

$\therefore {\left(3\right)}^{2}-3+k=0\phantom{\rule{0ex}{0ex}}⇒k+6=0\phantom{\rule{0ex}{0ex}}⇒k=-6$

The roots of the equation ${x}^{2}+2kx+\left({k}^{2}+2k+p\right)=0$ are equal.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(2k\right)}^{2}-4×1×\left({k}^{2}+2k+p\right)=0\phantom{\rule{0ex}{0ex}}⇒4{k}^{2}-4{k}^{2}-8k-4p=0\phantom{\rule{0ex}{0ex}}⇒-8k-4p=0$
$⇒p=\frac{8k}{-4}=-2k\phantom{\rule{0ex}{0ex}}⇒p=-2×\left(-6\right)=12$

Hence, the value of p is 12.

#### Page No 202:

It is given that −4 is a root of the quadratic equation ${x}^{2}+2x+4p=0$.

$\therefore {\left(-4\right)}^{2}+2×\left(-4\right)+4p=0\phantom{\rule{0ex}{0ex}}⇒16-8+4p=0\phantom{\rule{0ex}{0ex}}⇒4p+8=0\phantom{\rule{0ex}{0ex}}⇒p=-2$

The equation ${x}^{2}+px\left(1+3k\right)+7\left(3+2k\right)=0$ has equal roots.

$⇒4\left(1+6k+9{k}^{2}-21-14k\right)=0\phantom{\rule{0ex}{0ex}}⇒9{k}^{2}-8k-20=0\phantom{\rule{0ex}{0ex}}⇒9{k}^{2}-18k+10k-20=0\phantom{\rule{0ex}{0ex}}⇒9k\left(k-2\right)+10\left(k-2\right)=0$

Hence, the required value of k is 2 or $-\frac{10}{9}$.

#### Page No 203:

(i) The given equation is $k{x}^{2}+6x+1=0$.

$\therefore D={6}^{2}-4×k×1=36-4k$

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

$\therefore 36-4k>0\phantom{\rule{0ex}{0ex}}⇒4k<36\phantom{\rule{0ex}{0ex}}⇒k<9$

(ii) The given equation is ${x}^{2}-kx+9=0$.

$\therefore D={\left(-k\right)}^{2}-4×1×9={k}^{2}-36$

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

(iii) The given equation is $9{x}^{2}+3kx+4=0$.

$\therefore D={\left(3k\right)}^{2}-4×9×4=9{k}^{2}-144$

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

(iv) The given equation is $5{x}^{2}-kx+1=0$.

$\therefore D={\left(-k\right)}^{2}-4×5×1={k}^{2}-20$

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

#### Page No 203:

The given equation is $\left(a-b\right){x}^{2}+5\left(a+b\right)x-2\left(a-b\right)=0$.

Since a and b are real and ab, so ${\left(a-b\right)}^{2}>0$ and ${\left(a+b\right)}^{2}>0$.

$8{\left(a-b\right)}^{2}>0$    .....(1)               (Product of two positive numbers is always positive)

Also, $25{\left(a+b\right)}^{2}>0$             .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25{\left(a+b\right)}^{2}+8{\left(a-b\right)}^{2}>0$                 (Sum of two positive numbers is always positive)

$⇒D>0$

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

Ans

Ans

#### Page No 203:

It is given that the roots of the equation $a{x}^{2}+2bx+c=0$ are real.

Also, the roots of the equation $b{x}^{2}-2\sqrt{ac}x+b=0$ are real.
$\therefore {D}_{2}={\left(-2\sqrt{ac}\right)}^{2}-4×b×b\ge 0\phantom{\rule{0ex}{0ex}}⇒4\left(ac-{b}^{2}\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒-4\left({b}^{2}-ac\right)\ge 0$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
${b}^{2}-ac=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$

#### Page No 203:

Since,  $x=\frac{-1}{2}$ is a root of the quadratic equation 3x2 + 2kx + 3 = 0,
then, it must satisfies the equation.

#### Page No 203:

Since, $x=\frac{-1}{3}$ is a root of the quadratic equation 2x2 + 2x + k = 0,
then, it must satisfies the equation.

â€‹

#### Page No 203:

Given: x2 – 8x + 18 = 0

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

#### Page No 203:

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

It is given that, it has no real roots.

$⇒\mathrm{Discriminant}<0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac<0\phantom{\rule{0ex}{0ex}}⇒{\left(-12\right)}^{2}-4\left(4\right)\left(-k\right)<0\phantom{\rule{0ex}{0ex}}⇒144+16k<0\phantom{\rule{0ex}{0ex}}⇒16k<-144\phantom{\rule{0ex}{0ex}}⇒k<-9$

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

#### Page No 203:

Let one root be $\alpha$ and the other root be $\frac{1}{\alpha }$.

The given equation is 3x– 10k = 0.

Product of roots = $\frac{k}{3}$
$⇒\alpha ×\frac{1}{\alpha }=\frac{k}{3}\phantom{\rule{0ex}{0ex}}⇒1=\frac{k}{3}\phantom{\rule{0ex}{0ex}}⇒k=3$

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:

Let one root be $\alpha$ and the other root be $\frac{1}{\alpha }$.

The given equation is 5x2 +13k = 0.

Product of roots = $\frac{k}{5}$
$⇒\alpha ×\frac{1}{\alpha }=\frac{k}{5}\phantom{\rule{0ex}{0ex}}⇒1=\frac{k}{5}\phantom{\rule{0ex}{0ex}}⇒k=5$

Hence, the value of k is 5.

#### Page No 203:

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

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

$⇒\mathrm{Discriminant}=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}⇒{\left(k\right)}^{2}-4\left(3\right)\left(3\right)=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=36\phantom{\rule{0ex}{0ex}}⇒k=±6$

Hence, the values of k are –6 and 6.

#### Page No 203:

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

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

$⇒\mathrm{Discriminant}=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}⇒{\left(-3a\right)}^{2}-4\left(9\right)\left(1\right)=0\phantom{\rule{0ex}{0ex}}⇒9{a}^{2}=36\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=4\phantom{\rule{0ex}{0ex}}⇒a=±2$

Hence, the values of a are –2 and 2.

#### Page No 203:

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.

$⇒\mathrm{Discriminant}=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}⇒{\left(2k\right)}^{2}-4\left(1\right)\left({k}^{2}-k+2\right)=0\phantom{\rule{0ex}{0ex}}⇒4{k}^{2}-4{k}^{2}+4k-8=0\phantom{\rule{0ex}{0ex}}⇒4k=8\phantom{\rule{0ex}{0ex}}⇒k=2$

Hence, the value of k is 2.

#### Page No 224:

Let the required natural number be x.

According to the given condition,

$x+{x}^{2}=156\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-156=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+13x-12x-156=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+13\right)-12\left(x+13\right)=0$

x = 12     (x cannot be negative)

Hence, the required natural number is 12.

#### Page No 224:

Let the required natural number be x.

According to the given condition,

$x+\sqrt{x}=132$

Putting $\sqrt{x}=y$ or x = y2, we get

${y}^{2}+y=132\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+y-132=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+12y-11y-132=0\phantom{\rule{0ex}{0ex}}⇒y\left(y+12\right)-11\left(y+12\right)=0$

y = 11         (y cannot be negative)

Now,

$\sqrt{x}=11\phantom{\rule{0ex}{0ex}}⇒x={\left(11\right)}^{2}=121$

Hence, the required natural number is 121.

#### Page No 224:

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

According to the given condition,

$x\left(28-x\right)=192\phantom{\rule{0ex}{0ex}}⇒28x-{x}^{2}=192\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-28x+192=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x-12x+192=0$

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:

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

According to the given condition,
${x}^{2}+{\left(x+1\right)}^{2}=365\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+2x+1=365\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2x-364=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-182=0$

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:

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

According to the given condition,
${x}^{2}+{\left(x+2\right)}^{2}=514\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+4x+4=514\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x-510=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-255=0$

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:

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

According to the given condition,
${x}^{2}+{\left(x+2\right)}^{2}=452\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+4x+4=452\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x-448=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-224=0$

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:

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

According to the given condition,
$x\left(x+1\right)=306\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-306=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+18x-17x-306=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+18\right)-17\left(x+18\right)=0$

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:

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

According to the given condition,
$3x×3\left(x+1\right)=648\phantom{\rule{0ex}{0ex}}⇒9\left({x}^{2}+x\right)=648\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x=72\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-72=0$

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:

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

According to the given condition,
$x\left(x+2\right)=483\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-483=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+23x-21x-483=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+23\right)-21\left(x+23\right)=0$

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:

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

According to the given condition,
$x\left(x+2\right)=288\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-288=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+18x-16x-288=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+18\right)-16\left(x+18\right)=0$

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:

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

According to the given condition,
$\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{9-x+x}{x\left(9-x\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{9}{9x-{x}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒9x-{x}^{2}=18$
$⇒{x}^{2}-9x+18=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x-3x+18=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-6\right)-3\left(x-6\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-3\right)\left(x-6\right)=0$

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:

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

According to the given condition,
$\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒\frac{15-x+x}{x\left(15-x\right)}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒\frac{15}{15x-{x}^{2}}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒15x-{x}^{2}=50$
$⇒{x}^{2}-15x+50=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-10x-5x+50=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-10\right)-5\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-5\right)\left(x-10\right)=0$

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:

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

Now, x < x + 3
$\therefore \frac{1}{x}>\frac{1}{x+3}$

According to the given condition,
$\frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒\frac{x+3-x}{x\left(x+3\right)}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{{x}^{2}+3x}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+3x=28$
$⇒{x}^{2}+3x-28=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+7x-4x-28=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+7\right)-4\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+7\right)\left(x-4\right)=0$

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:

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

Now, x < x + 5
$\therefore \frac{1}{x}>\frac{1}{x+5}$

According to the given condition,
$\frac{1}{x}-\frac{1}{x+5}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{{x}^{2}+5x}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5x=14$
$⇒{x}^{2}+5x-14=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+7x-2x-14=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+7\right)-2\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+7\right)\left(x-2\right)=0$

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:

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

According to the given condition,
${\left(7x\right)}^{2}+{\left[7\left(x+1\right)\right]}^{2}=1225\phantom{\rule{0ex}{0ex}}⇒49{x}^{2}+49\left({x}^{2}+2x+1\right)=1225\phantom{\rule{0ex}{0ex}}⇒49{x}^{2}+49{x}^{2}+98x+49=1225\phantom{\rule{0ex}{0ex}}⇒98{x}^{2}+98x-1176=0$
$⇒{x}^{2}+x-12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-3x-12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+4\right)-3\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+4\right)\left(x-3\right)=0$

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:

Let the natural number be x.

According to the given condition,
$x+\frac{1}{x}=\frac{65}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+1}{x}=\frac{65}{8}\phantom{\rule{0ex}{0ex}}⇒8{x}^{2}+8=65x\phantom{\rule{0ex}{0ex}}⇒8{x}^{2}-65x+8=0$

x = 8         (x is a natural number)

Hence, the required number is 8.

#### Page No 225:

Let the two parts be x and (57 − x).

According to the given condition,
$x\left(57-x\right)=680\phantom{\rule{0ex}{0ex}}⇒57x-{x}^{2}=680\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-57x+680=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-40x-17x+680=0$

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:

Let the two parts be x and (27 − x).

According to the given condition,
$\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{27-x+x}{x\left(27-x\right)}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{27}{27x-{x}^{2}}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒27x-{x}^{2}=180$
$⇒{x}^{2}-27x+180=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-15x-12x+180=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-15\right)-12\left(x-15\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-12\right)\left(x-15\right)=0$

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:

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

According to the given condition,
${x}^{2}+\left(x+1\right)\left(x+2\right)=46\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+3x+2=46\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+3x-44=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+11x-8x-44=0$

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.

â€‹â€‹

#### Page No 225:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

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

According to the given condition,
$\frac{x-3}{x+1}=\frac{x-3}{x}-\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x-3\right)\left(x+1\right)-x\left(x-3\right)}{x\left(x+1\right)}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}-2x-3-{x}^{2}+3x}{{x}^{2}+x}=\frac{1}{15}$
$⇒\frac{x-3}{{x}^{2}+x}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x=15x-45\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+45=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-9x-5x+45=0$

When x = 5,
$\frac{x-3}{x}=\frac{5-3}{5}=\frac{2}{5}$

When x = 9,
$\frac{x-3}{x}=\frac{9-3}{9}=\frac{6}{9}=\frac{2}{3}$             (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$.

#### Page No 225:

Let the required number be x.

According to the given condition,
$x+\frac{1}{x}=2\frac{1}{30}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+1}{x}=\frac{61}{30}\phantom{\rule{0ex}{0ex}}⇒30{x}^{2}+30=61x\phantom{\rule{0ex}{0ex}}⇒30{x}^{2}-61x+30=0$

Hence, the required number is $\frac{5}{6}$ or $\frac{6}{5}$.

#### Page No 226:

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

∴ Original cost of food for each member = â‚ą$\frac{2000}{x}$

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

∴ New cost of food for each member = â‚ą$\frac{2000}{\left(x-5\right)}$

According to the given condition,
â‚ą$\frac{2000}{x-5}$ − â‚ą$\frac{2000}{x}$ = â‚ą20
$⇒\frac{2000x-2000x+10000}{x\left(x-5\right)}=20\phantom{\rule{0ex}{0ex}}⇒\frac{10000}{{x}^{2}-5x}=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x=500\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x-500=0$

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 = â‚ą$\frac{2000}{\left(25-5\right)}$ = â‚ą$\frac{2000}{20}$ = â‚ą100

#### Page No 226:

Let the original price of the book be â‚ąx.

∴ Number of books bought at original price for â‚ą600 = $\frac{600}{x}$

If the price of a book is reduced by â‚ą5, then the new price of the book is â‚ą(x − 5).

∴ Number of books bought at reduced price for â‚ą600 = $\frac{600}{x-5}$

According to the given condition,
$\frac{600}{x-5}-\frac{600}{x}=4\phantom{\rule{0ex}{0ex}}⇒\frac{600x-600x+3000}{x\left(x-5\right)}=4\phantom{\rule{0ex}{0ex}}⇒\frac{3000}{{x}^{2}-5x}=4\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x=750$
$⇒{x}^{2}-5x-750=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-30x+25x-750=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-30\right)+25\left(x-30\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-30\right)\left(x+25\right)=0$

x = 30               (Price cannot be negative)

Hence, the original price of the book is â‚ą30.

#### Page No 226:

Let the original duration of the tour be x days.

∴ Original daily expenses = â‚ą$\frac{10,800}{x}$

If he extends his tour by 4 days, then his new daily expenses = â‚ą$\frac{10,800}{x+4}$

According to the given condition,

â‚ą$\frac{10,800}{x}$ − â‚ą$\frac{10,800}{x+4}$ = â‚ą90

$⇒\frac{10800x+43200-10800x}{x\left(x+4\right)}=90\phantom{\rule{0ex}{0ex}}⇒\frac{43200}{{x}^{2}+4x}=90\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x=480\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-480=0$

x = 20              (Number of days cannot be negative)

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

#### Page No 226:

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

According to the given condition,
$\left(x+3\right)\left(28-x-4\right)=180\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)\left(24-x\right)=180\phantom{\rule{0ex}{0ex}}⇒-{x}^{2}+21x+72=180\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-21x+108=0$

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:

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

Also, given that if 3 more pens are purchased for the same amount, the price per pen gets reduced by Rs 3

Put the value of x in (ii) we get

We ignore the negative value.
$⇒x=12$
Therefore, number of pens is 12.

#### Page No 226:

Let the cost price of the article be â‚ąx.

∴ Gain percent = x%

According to the given condition,
â‚ąx + â‚ą$\left(\frac{x}{100}×x\right)$ = â‚ą75                            (Cost price + Gain = Selling price)
$⇒\frac{100x+{x}^{2}}{100}=75\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x=7500\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x-7500=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+150x-50x-7500=0$

x = 50                   (Cost price cannot be negative)

Hence, the cost price of the article is â‚ą50.

#### Page No 226:

(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

$\therefore {x}^{2}-1=8\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-1=8x-8\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x+7=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x-x+7=0$

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 $3\frac{1}{2}$ times as old as his son.

Also given that

Put value of (i) in (ii), we get
${\left(\frac{7}{2}s\right)}^{2}+{s}^{2}=1325\phantom{\rule{0ex}{0ex}}\frac{⇒49{s}^{2}+4{s}^{2}}{4}=1325\phantom{\rule{0ex}{0ex}}53{s}^{2}=5300\phantom{\rule{0ex}{0ex}}⇒{s}^{2}=100\phantom{\rule{0ex}{0ex}}⇒s=±10$
Ignore the negative value
So, the age of son = s = 10 years
Also, from (i) we have
$m=\frac{7}{2}s\phantom{\rule{0ex}{0ex}}⇒m=\frac{7}{2}×10\phantom{\rule{0ex}{0ex}}⇒m=35$
So, age of man = 35 years
Age of son = 10 years

#### Page No 226:

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,
$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5+x-3}{\left(x-3\right)\left(x+5\right)}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+2}{{x}^{2}+2x-15}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-15=6x+6$
$⇒{x}^{2}-4x-21=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x+3x-21=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-7\right)+3\left(x-7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-7\right)\left(x+3\right)=0$

x = 7                 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

#### Page No 227:

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)

$⇒3{x}^{2}+5=4x+20\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-4x-15=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-9x+5x-15=0\phantom{\rule{0ex}{0ex}}⇒3x\left(x-3\right)+5\left(x-3\right)=0$

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:

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

∴ Time taken to cover 150 km = $\frac{150}{x}$ h

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

∴ Time taken to cover 200 km = $\frac{200}{x+20}$ h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
$\therefore \frac{150}{x}+\frac{200}{x+20}=5\phantom{\rule{0ex}{0ex}}⇒\frac{150x+3000+200x}{x\left(x+20\right)}=5\phantom{\rule{0ex}{0ex}}⇒350x+3000=5\left({x}^{2}+20x\right)\phantom{\rule{0ex}{0ex}}⇒350x+3000=5{x}^{2}+100x$
$⇒5{x}^{2}-250x-3000=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-50x-600=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-60x+10x-600=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-60\right)+10\left(x-60\right)=0$

x = 60                  (Speed cannot be negative)

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

#### Page No 227:

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 = $\frac{1500}{x}$ h                             $\left(\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\right)$

Time taken to reach the destination at actual speed = $\frac{1500}{x+100}$ 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

$⇒{x}^{2}+100x=300000\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x-300000=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+600x-500x-300000=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+600\right)-500\left(x+600\right)=0$

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:

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 = $\frac{480}{x}$ h                  $\left(\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\right)$

Time taken by the train to cover the distance at reduced speed = $\frac{480}{x-8}$ 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

$\therefore \frac{480}{x-8}=\frac{480}{x}+3\phantom{\rule{0ex}{0ex}}⇒\frac{480}{x-8}-\frac{480}{x}=3\phantom{\rule{0ex}{0ex}}⇒\frac{480x-480x+3840}{x\left(x-8\right)}=3\phantom{\rule{0ex}{0ex}}⇒\frac{3840}{{x}^{2}-8x}=3$
$⇒{x}^{2}-8x=1280\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x-1280=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-40x+32x-1280=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-40\right)+32\left(x-40\right)=0$

x = 40              (Speed cannot be negative)

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

#### Page No 227:

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

∴ Time taken to cover 54 km = $\frac{54}{x}$ h

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

∴ Time taken to cover 63 km = $\frac{63}{x+6}$ h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
$\therefore \frac{54}{x}+\frac{63}{x+6}=3\phantom{\rule{0ex}{0ex}}⇒\frac{54x+324+63x}{x\left(x+6\right)}=3\phantom{\rule{0ex}{0ex}}⇒117x+324=3\left({x}^{2}+6x\right)\phantom{\rule{0ex}{0ex}}⇒117x+324=3{x}^{2}+18x$
$⇒3{x}^{2}-99x-324=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-33x-108=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-36x+3x-108=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-36\right)+3\left(x-36\right)=0$

x = 36                  (Speed cannot be negative)

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

#### Page No 227:

Distance covered by the train = 180 km
We know that distance covered$\left(d\right)=\mathrm{speed}\left(s\right)×\mathrm{time}\left(t\right)$

Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour.

Put the value of (i) in (ii), we get

Ignore the negative value
So, time taken = 5 hours
From (i)
$s=\frac{180}{t}=\frac{180}{5}=36$
Hence, the speed is 36 km/h.

#### Page No 227:

Distance covered by the train = 300 km
We know that distance covered$\left(d\right)=\mathrm{speed}\left(s\right)×\mathrm{time}\left(t\right)$

Also, given that if the speed is increased by 5 km/h, time of travel gets reduced by 2 hours.

Put the value of (i) in (ii), we get

Ignore the negative value
Therefore, the original speed = 25 km/h

#### Page No 228:

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
Total speed of the boat while going upstream $=15-x$ km/h
Time taken to go upstream $=\frac{30}{15-x}$ hrs
Total speed of boat while going downstream $=15+x$ km/h
Time taken to go downstream $=\frac{30}{15+x}$ hrs
Total time of the journey $=4\frac{1}{2}$hrs = 4.5 hrs
$⇒\frac{30}{15-x}+\frac{30}{15+x}=4.5\phantom{\rule{0ex}{0ex}}⇒\frac{30\left[\left(15+x\right)+\left(15-x\right)\right]}{\left(15{\right)}^{2}-{x}^{2}}=4.5\phantom{\rule{0ex}{0ex}}⇒30\left[15+x+15-x\right]=4.5\left[\left(15{\right)}^{2}-{x}^{2}\right]\phantom{\rule{0ex}{0ex}}⇒30\left[30\right]=4.5\left[225-{x}^{2}\right]\phantom{\rule{0ex}{0ex}}⇒\frac{30×30}{4.5}=225-{x}^{2}\phantom{\rule{0ex}{0ex}}⇒225-{x}^{2}=200\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=25\phantom{\rule{0ex}{0ex}}⇒x=±5$
Ignore the negative value.
So, the speed of the stream = x = 5 km/h

#### Page No 228:

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 = $\frac{V}{x}$

⇒ Volume of the tank filled by one pipe in $11\frac{1}{9}$ minutes = $\frac{V}{x}×11\frac{1}{9}=\frac{V}{x}×\frac{100}{9}$

Similarly,

Volume of the tank filled by the other pipe in $11\frac{1}{9}$ minutes = $\frac{V}{\left(x+5\right)}×11\frac{1}{9}=\frac{V}{\left(x+5\right)}×\frac{100}{9}$

Now,

Volume of the tank filled by one pipe in $11\frac{1}{9}$ minutes + Volume of the tank filled by the other pipe in $11\frac{1}{9}$ minutes = V
$\therefore V\left(\frac{1}{x}+\frac{1}{x+5}\right)×\frac{100}{9}=V\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5+x}{x\left(x+5\right)}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+5}{{x}^{2}+5x}=\frac{9}{100}$
$⇒200x+500=9{x}^{2}+45x\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-155x-500=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-180x+25x-500=0\phantom{\rule{0ex}{0ex}}⇒9x\left(x-20\right)+25\left(x-20\right)=0$

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:

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 = $\frac{V}{x}$

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = $\frac{V}{x}×6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = $\frac{V}{\left(x-9\right)}×6$

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
$\therefore V\left(\frac{1}{x}+\frac{1}{x-9}\right)×6=V\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{x-9}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{x-9+x}{x\left(x-9\right)}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-9}{{x}^{2}-9x}=\frac{1}{6}$
$⇒12x-54={x}^{2}-9x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-21x+54=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-18x-3x+54=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-18\right)-3\left(x-18\right)=0$

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:

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

Let the length of rectangular plot be x m.
$\mathrm{Perimeter}=60\phantom{\rule{0ex}{0ex}}⇒2\left(\mathrm{Length}+\mathrm{Breadth}\right)=60\phantom{\rule{0ex}{0ex}}⇒\mathrm{Length}+\mathrm{Breadth}=30\phantom{\rule{0ex}{0ex}}⇒\mathrm{Breadth}=30-\mathrm{Length}$

Then, the breadth is (30 − x) m.

According to the question,

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

#### Page No 229:

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

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

#### Page No 229:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (+ 10) cm.

#### Page No 229:

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

Area of a triangle =

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:

Let the base be $x$ m.
Therefore, the altitude will be m.

Area of a triangle =

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

#### Page No 229:

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

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:

Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively.
Therefore, the hypotenuse will be  cm.

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

#### Page No 229:

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

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

#### Page No 229:

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.

$\frac{\mathrm{Distance}}{\mathrm{Speed}}=\mathrm{Time}$

Time taken by faster train to cover 200 km = $\frac{200}{x}$ h
Time taken by slower train to cover 200 km = $\frac{200}{x-10}$ h

According to the question,

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:

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  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

$\frac{\mathrm{Distance}}{\mathrm{Speed}}=\mathrm{Time}$

According to the question,

Time to go upstream = $\frac{3}{2}$ hours + Time to go downstream

Time to go upstream − Time to go downstream = $\frac{3}{2}$

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

#### Page No 229:

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,

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

#### Page No 229:

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 = $\frac{1}{x}$
Tank filled in 1 hour by larger tap = $\frac{1}{x-2}$
Tank filled in 1 hour by both the taps = $\frac{8}{15}$

According to the question,

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

(d)

(b)

(c)

(b) −11

(b) −7

(c) 6

(c)  8

(c) 2 : 3

(d) 3

(d) 5

#### Page No 238:

(d) $\frac{-2}{3}$

(b)

(a)

#### Page No 238:

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

and $\alpha \beta =\frac{2}{3}$
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{8}{3}}{\frac{2}{3}}=-4$

Hence, the correct answer is option C.

(c) c  =  a

#### Page No 238:

(d) $\frac{{b}^{2}}{4a}$

(c) 2 or −2

(a) 1 or 4

#### Page No 239:

(d) $±\frac{4}{3}$

(a) >  0

#### Page No 239:

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.

(d) imaginary

#### Page No 239:

(b) real, unequal and irrational

(d) either or

(c)

(c) −2  <   <  2

(b)

(a)

(c) 16 m

#### Page No 240:

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)

$\therefore \left(x+8\right)×x=240$                       (Area = Length × Breadth)
$⇒{x}^{2}+8x-240=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+20x-12x-240=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+20\right)-12\left(x+20\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+20\right)\left(x-12\right)=0$

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:

The given quadratic equation is $2{x}^{2}-x-6=0$.

$2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(2x+3\right)=0$

Thus, the roots of the given equation are 2 and $\frac{-3}{2}$.

Hence, the correct answer is option B.

#### Page No 240:

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

It is given that the product of the two numbers is 15.
$\therefore x\left(8-x\right)=15\phantom{\rule{0ex}{0ex}}⇒8x-{x}^{2}=15\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x+15=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x-3x+15=0$

Hence, the required numbers are 3 and 5.

#### Page No 240:

The given equation is ${x}^{2}+6x+9=0$.

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

LHS = ${\left(-3\right)}^{2}+6×\left(-3\right)+9=9-18+9=0$ = RHS

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

#### Page No 240:

The given equation is $3{x}^{2}+13x+14=0$.

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

LHS = $3×{\left(-2\right)}^{2}+13×\left(-2\right)+14=12-26+14=0$ = RHS

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

#### Page No 240:

It is given that $x=\frac{-1}{2}$ is a solution of the quadratic equation $3{x}^{2}+2kx-3=0$.

$\therefore 3×{\left(\frac{-1}{2}\right)}^{2}+2k×\left(\frac{-1}{2}\right)-3=0\phantom{\rule{0ex}{0ex}}⇒\frac{3}{4}-k-3=0\phantom{\rule{0ex}{0ex}}⇒k=\frac{3-12}{4}=-\frac{9}{4}$

Hence, the value of k is $-\frac{9}{4}$.

#### Page No 240:

The given quadratic equation is $2{x}^{2}-x-6=0$.
$2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(2x+3\right)=0$

Hence, the roots of the given equation are 2 and $-\frac{3}{2}$.

#### Page No 240:

The given quadratic equation is $3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0$.
$3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{3}{x}^{2}+9x+x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{3}x\left(x+\sqrt{3}\right)+1\left(x+\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+\sqrt{3}\right)\left(3\sqrt{3}x+1\right)=0$

Hence, $-\sqrt{3}$ and $-\frac{\sqrt{3}}{9}$ are the solutions of the given equation.

#### Page No 240:

It is given that the roots of the quadratic equation $2{x}^{2}+8x+k=0$ are equal.
$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{8}^{2}-4×2×k=0\phantom{\rule{0ex}{0ex}}⇒64-8k=0\phantom{\rule{0ex}{0ex}}⇒k=8$
Hence, the value of k is 8.

#### Page No 240:

It is given that the quadratic equation $p{x}^{2}-2\sqrt{5}px+15=0$ has two equal roots.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(-2\sqrt{5}p\right)}^{2}-4×p×15=0\phantom{\rule{0ex}{0ex}}⇒20{p}^{2}-60p=0\phantom{\rule{0ex}{0ex}}⇒20p\left(p-3\right)=0$

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

p ≠ 0

Hence, the value of p is 3.

#### Page No 240:

It is given that y = 1 is a root of the equation $a{y}^{2}+ay+3=0$.
$\therefore a×{\left(1\right)}^{2}+a×1+3=0\phantom{\rule{0ex}{0ex}}⇒a+a+3=0\phantom{\rule{0ex}{0ex}}⇒2a+3=0\phantom{\rule{0ex}{0ex}}⇒a=-\frac{3}{2}$

Also, y = 1 is a root of the equation ${y}^{2}+y+b=0$.
$\therefore {\left(1\right)}^{2}+1+b=0\phantom{\rule{0ex}{0ex}}⇒1+1+b=0\phantom{\rule{0ex}{0ex}}⇒b+2=0\phantom{\rule{0ex}{0ex}}⇒b=-2$

$\therefore ab=\left(-\frac{3}{2}\right)×\left(-2\right)=3$

Hence, the value of ab is 3.

#### Page No 240:

Let the other zero of the given polynomial be $\alpha$.

Now,
Sum of the zeroes of the given polynomial = $\frac{-\left(-4\right)}{1}$= 4

Hence, the other zero of the given polynomial is $\left(2-\sqrt{3}\right)$.

#### Page No 240:

Let $\alpha$ and $\beta$ be the roots of the equation $3{x}^{2}-10x+k=0$.
(Given)

Hence, the value of k is 3.

#### Page No 240:

It is given that the roots of the quadratic equation $p{x}^{2}-2px+6=0$ are equal.
$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(-2p\right)}^{2}-4×p×6=0\phantom{\rule{0ex}{0ex}}⇒4{p}^{2}-24p=0\phantom{\rule{0ex}{0ex}}⇒4p\left(p-6\right)=0$

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

p ≠ 0

Hence, the value of p is 6.

#### Page No 240:

It is given that the quadratic equation ${x}^{2}-4kx+k=0$ has equal roots.

Hence, 0 and $\frac{1}{4}$ are the required values of k.

#### Page No 240:

It is given that the quadratic equation $9{x}^{2}-3kx+k=0$ has equal roots.

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

#### Page No 240:

${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\sqrt{3}x-x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-\sqrt{3}\right)-1\left(x-\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-\sqrt{3}\right)\left(x-1\right)=0$

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

#### Page No 240:

$2{x}^{2}+ax-{a}^{2}=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2ax-ax-{a}^{2}=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x+a\right)-a\left(x+a\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+a\right)\left(2x-a\right)=0$

Hence, −a and $\frac{a}{2}$ are the roots of the given equation.

#### Page No 241:

$3{x}^{2}+5\sqrt{5}x-10=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}+6\sqrt{5}x-\sqrt{5}x-10=0\phantom{\rule{0ex}{0ex}}⇒3x\left(x+2\sqrt{5}\right)-\sqrt{5}\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+2\sqrt{5}\right)\left(3x-\sqrt{5}\right)=0$

Hence, $-2\sqrt{5}$ and $\frac{\sqrt{5}}{3}$ are the roots of the given equation.

#### Page No 241:

$\sqrt{3}{x}^{2}+10x-8\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}{x}^{2}+12x-2x-8\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(x+4\sqrt{3}\right)-2\left(x+4\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+4\sqrt{3}\right)\left(\sqrt{3}x-2\right)=0$

Hence, $-4\sqrt{3}$ and $\frac{2\sqrt{3}}{3}$ are the roots of the given equation.

#### Page No 241:

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}{x}^{2}-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(x-\sqrt{6}\right)+\sqrt{2}\left(x-\sqrt{6}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-\sqrt{6}\right)\left(\sqrt{3}x+\sqrt{2}\right)=0$

Hence, $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$ are the roots of the given equation.

#### Page No 241:

$4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4\sqrt{3}{x}^{2}+8x-3x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒4x\left(\sqrt{3}x+2\right)-\sqrt{3}\left(\sqrt{3}x+2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}x+2\right)\left(4x-\sqrt{3}\right)=0$

Hence, $-\frac{2\sqrt{3}}{3}$ and $\frac{\sqrt{3}}{4}$ are the roots of the given equation.

#### Page No 241:

$4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+4bx-\left(a-b\right)\left(a+b\right)=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+2\left[\left(a+b\right)-\left(a-b\right)\right]x-\left(a-b\right)\left(a+b\right)=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+2\left(a+b\right)x-2\left(a-b\right)x-\left(a-b\right)\left(a+b\right)=0$

Hence, $-\frac{a+b}{2}$ and $\frac{a-b}{2}$ are the roots of the given equation.

#### Page No 241:

${x}^{2}+5x-\left({a}^{2}+a-6\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5x-\left(a+3\right)\left(a-2\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\left[\left(a+3\right)-\left(a-2\right)\right]x-\left(a+3\right)\left(a-2\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\left(a+3\right)x-\left(a-2\right)x-\left(a+3\right)\left(a-2\right)=0$

Hence, $-\left(a+3\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Page No 241:

${x}^{2}+6x-\left({a}^{2}+2a-8\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+6x-\left(a+4\right)\left(a-2\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\left[\left(a+4\right)-\left(a-2\right)\right]x-\left(a+4\right)\left(a-2\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\left(a+4\right)x-\left(a-2\right)x-\left(a+4\right)\left(a-2\right)=0$

Hence, $-\left(a+4\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Page No 241:

${x}^{2}-4ax+4{a}^{2}-{b}^{2}=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4ax+\left(2a+b\right)\left(2a-b\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\left[\left(2a+b\right)+\left(2a-b\right)\right]x+\left(2a+b\right)\left(2a-b\right)=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\left(2a+b\right)x-\left(2a-b\right)x+\left(2a+b\right)\left(2a-b\right)=0$