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#### Question 1:

(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.

#### Question 2:

(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.

#### Question 3:

(i)

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

Hence, the other root is 3.

(ii)

#### Question 4:

(i)

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

Hence, the other root is 3.

(ii)

LHS;
$a{d}^{2}\left(\frac{ax}{b}+\frac{2c}{d}\right)x+b{c}^{2}=0$
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.

#### Question 5:

LHS;
$a{d}^{2}\left(\frac{ax}{b}+\frac{2c}{d}\right)x+b{c}^{2}=0$
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.

(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 $-\frac{1}{3}$.

#### Question 6:

(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 $-\frac{1}{3}$.

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 $-\frac{5}{4}$.

#### Question 7:

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 $-\frac{5}{4}$.

#### Question 8:

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 $\frac{3}{2}$.

#### Question 9:

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 $\frac{3}{2}$.

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.

#### Question 10:

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.

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 $\frac{2}{3}$.

#### Question 11:

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 $\frac{2}{3}$.

#### Question 15:

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 $-\frac{1}{3}$.

#### Question 16:

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 $-\frac{1}{3}$.

#### Question 20:

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 $\sqrt{2}$.

#### Question 21:

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 $\sqrt{2}$.

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 $-4\sqrt{3}$.

#### Question 22:

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 $-4\sqrt{3}$.

#### Question 24:

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 $\frac{13\sqrt{7}}{7}$.

#### Question 25:

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 $\frac{13\sqrt{7}}{7}$.

#### Question 26:

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.

#### Question 27:

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.

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 $-\frac{\sqrt{6}}{3}$.

#### Question 28:

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 $-\frac{\sqrt{6}}{3}$.

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 $2\sqrt{5}$.

#### Question 29:

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 $2\sqrt{5}$.

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

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

#### Question 30:

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

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

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 $2\sqrt{3}$.

#### Question 31:

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 $2\sqrt{3}$.

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 $-\frac{5\sqrt{2}}{2}$.

#### Question 32:

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 $-\frac{5\sqrt{2}}{2}$.

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.

#### Question 33:

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.

#### Question 35:

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.

#### Question 36:

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.

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.

#### Question 37:

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.

#### Question 39:

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.

#### Question 40:

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.

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.

#### Question 41:

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.

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.

#### Question 42:

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.

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.

#### Question 43:

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.

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.

#### Question 44:

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.

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.

#### Question 45:

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.

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.

#### Question 46:

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.

#### Question 47:

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.

#### Question 48:

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.

#### Question 51:

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.

#### Question 52:

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.

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

#### Question 53:

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

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

#### Question 54:

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

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

#### Question 55:

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

(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)

#### Question 56:

(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)

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

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

#### Question 57:

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

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

#### Question 58:

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

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

#### Question 59:

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

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

(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)

#### Question 60:

(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)

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

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

#### Question 61:

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

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

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

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

#### Question 62:

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

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

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

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

#### Question 63:

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

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

#### Question 64:

(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)

#### Question 65:

(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)

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

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

#### Question 66:

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

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

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

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

#### Question 67:

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

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

#### Question 1:

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€‹

#### Question 2:

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€‹

#### Question 4:

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.

#### Question 5:

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.

#### Question 8:

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.

#### Question 9:

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.

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.

#### Question 10:

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.

#### Question 11:

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.

#### Question 12:

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.

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.

#### Question 13:

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.

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.

#### Question 14:

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.

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.

#### Question 15:

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.

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.

#### Question 16:

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.

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).

#### Question 17:

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).

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.

#### Question 18:

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.

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.

#### Question 19:

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.

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.

#### Question 20:

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.

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).

#### Question 21:

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).

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.

#### Question 22:

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.

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.

#### Question 23:

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.

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.

#### Question 24:

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.

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.

#### Question 25:

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.

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.

#### Question 26:

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.

#### Question 27:

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.

#### Question 28:

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.

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.

#### Question 29:

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.

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.

#### Question 30:

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.

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.

#### Question 31:

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.

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.

#### Question 32:

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.

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.

#### Question 33:

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.

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.

#### Question 34:

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.

#### Question 35:

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.

#### Question 36:

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.

#### Question 1:

(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.

#### Question 2:

(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.

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.

#### Question 3:

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.

#### Question 5:

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.

#### Question 6:

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.

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.

#### Question 7:

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.

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.

#### Question 8:

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.

(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)

#### Question 9:

(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)

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.

#### Question 10:

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.

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.

#### Question 11:

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.

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 $\frac{7}{4}$.

#### Question 12:

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 $\frac{7}{4}$.

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.

#### Question 13:

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.

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 $-\frac{10}{9}$.

#### Question 14:

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 $-\frac{10}{9}$.

#### Question 19:

(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.

#### Question 20:

(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.

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.

#### Question 21:

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

Ans

#### Question 23:

Ans

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
${b}^{2}-ac=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$

#### Question 24:

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
${b}^{2}-ac=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$

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

#### Question 25:

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

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

â€‹

#### Question 26:

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

â€‹

Given: x2 – 8x + 18 = 0

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

#### Question 27:

Given: x2 – 8x + 18 = 0

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

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.

#### Question 28:

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.

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.

#### Question 29:

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.

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.

#### Question 30:

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.

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.

#### Question 31:

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.

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.

#### Question 32:

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.

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.

#### Question 1:

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.

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.

#### Question 2:

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.

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.

#### Question 3:

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.

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.

#### Question 4:

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.

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.