Thinking Process

An equation which is of the form $a{x}^2+bx+c=0a\ne 0$ is called a quadratic equation. So, simplify each part of the question and check whether it is in the form of $a{x}^2+bx+c=0,a\ne 0$ or not.

Solution

(d)

(a) Given that, $\begin{array}{ll}& x^2+2x+1=(4-x{)}^2+3\\ \Rightarrow & x^2+2x+1=16+x^2-8x+3\\ \Rightarrow & 10x-18=0\end{array}$

which is not of the form $a{x}^2+bx+c,a\ne 0$. Thus, the equation is not a quadratic

(b) Given that,

$\begin{array}{ccc}& -2x^2& =(5-x)2x-{\displaystyle \frac{2}{5}}\\ \Rightarrow & & -2x^2& =10x-2x^2-2+{\displaystyle \frac{2x}{5}}\\ \Rightarrow & & 50x+2x-10& =0\\ \Rightarrow & & 52x-10& =0\end{array}$

(c) Given that,

$\begin{array}{rlrl}& & x^2(k+1)+{\displaystyle \frac{3}{2}}x& =7\\ \text{ Given }k& & =-1\\ \Rightarrow & x^2(-1+1)+{\displaystyle \frac{3}{2}}x& =7\\ \Rightarrow & 3x-14& =0\end{array}$

which is also not a quadratic equation.

(d)

$\begin{array}{ccc}\text{ Given that, }& x^3-x^2& =(x-1{)}^3\\ \Rightarrow & x^3-x^2& =x^3-3x^2(1)+3x(1{)}^2-(1{)}^3\\ & [\therefore (a-b{)}^3& =a^3-b^3+3a{b}^2-3a^{2}b]\\ \Rightarrow & x^3-x^2& =x^3-3x^2+3x-1\\ \Rightarrow & -x^2+3x^2-3x+1& =0\\ \Rightarrow & 2x^2-3x+1& =0\end{array}$

$\text{ which represents a quadratic equation because it has the quadratic form }$

$a{x}^2+bx+c=0,a\ne 0$.

Solution

(d)

(a)

$\begin{array}{ccc}\text{ Given that, }& 2(x-1{)}^2& =4x^2-2x+1\\ \Rightarrow & 2(x^2+1-2x)& =4x^2-2x+1\\ \Rightarrow & 2x^2+2-4x& =4x^2-2x+1\\ \Rightarrow & 2x^2+2x-1& =0\end{array}$

which represents a quadratic equation because it has the quadratic form $a{x}^2+bx+c=0,a\ne 0$.

(b)

$\begin{array}{rlr}\text{ Given that, }& 2x-x^2& =x^2+5\\ \Rightarrow & & 2x^2-2x+5& =0\end{array}$

which also represents a quadratic equation because it has the quadratic form $a{x}^2+bx+c=0,a\ne 0$.

(c)

$\begin{array}{rlrl}\text{ Given that, }& & (\sqrt{2}\cdot x+\sqrt{3}{)}^2& =3x^2-5x\\ \Rightarrow & & 2\cdot x^2+3+2\sqrt{6}\cdot x& =3x^2-5x\\ \Rightarrow & & x^2-(5+2\sqrt{6})x-3& =0\end{array}$

which also represents a quadratic equation because it has the quadratic form $a{x}^2+bx+c=0,a\ne 0$.

(d) Given that,

$\begin{array}{rlrl}\text{ Given that, }& & (x^2+2x{)}^2& =x^4+3+4x^2\\ \Rightarrow & & -4x^2+4x^3& =x^4+3+4x^2\\ \Rightarrow & & 4x^3-3& =0\end{array}$

which is not of the form $a{x}^2+bx+c,a\ne 0$. Thus, the equation is not quadratic. This is a cubic equation.

Thinking Process

If $\mathit{\alpha }$ is one of the root of any quadratic equation i.e., $f(x)=a{x}^2+bx+c=0$, then $x=\alpha$ i.e., $\mathit{\alpha }$ satisfies the equation $a{\mathit{\alpha }}^2+b\mathit{\alpha }+c=0$.

Solution

(c)

(a) Substituting $x=2$ in $x^2-4x+5$, we get (2) ${)}^2-4$ (2) +5

$$=4-8+5=1\ne 0$$

So, $x=2$ is not a root of $x^2-4x+5=0$.

(b) Substituting $x=2$ in $x^2+3x-12$, we get

$$\begin{array}{rl}& (2{)}^2+3(2)-12\\ & =4+6-12=-2\ne 0\end{array}$$

So, $x=2$ is not a root of $x^2+3x-12=0$.

(c) Substituting $x=2$ in $2x^2-7x+6$, we get

$$\begin{array}{rl}2(2{)}^2-7(2)+6& =2(4)-14+6\\ & =8-14+6=14-14=0\end{array}$$

So, $x=2$ is root of the equation $2x^2-7x+6=0$.

(d) Substituting $x=2$ in $3x^2-6x-2$, we get

$$\begin{array}{rl}& 3(2{)}^2-6(2)-2\\ & =12-12-2=-2\ne 0\end{array}$$

So, $x=2$ is not a root of $3x^2-6x-2=0$.

Thinking Process

Since, ${\displaystyle \frac{1}{2}}$ is a root of given equation, then put $x={\displaystyle \frac{1}{2}}$ in given equation and get the value of $k$.

Solution

(a) Since, ${\displaystyle \frac{1}{2}}$ is a root of the quadratic equation $x^2+kx-{\displaystyle \frac{5}{4}}=0$.

$\begin{array}{ccc}\text{ Then, }& & ({\displaystyle \frac{1}{2}}{)}^2+k({\displaystyle \frac{1}{2}})-{\displaystyle \frac{5}{4}}=0\\ \Rightarrow & & {\displaystyle \frac{1}{4}}+{\displaystyle \frac{k}{2}}-{\displaystyle \frac{5}{4}}=0\Rightarrow {\displaystyle \frac{1+2k-5}{4}}=0\\ \Rightarrow & & 2k-4=0\\ \Rightarrow & & 2k=4\Rightarrow k=2\end{array}$

Thinking Process

If $\mathit{\alpha }$ and $\mathit{\beta }$ are the roots of the quadratic equation $a{x}^2+bx+c=0,a\ne 0$, then sum of roots $=\alpha +\beta =(-1){\displaystyle \frac{\text{ Coefficient of }x}{\text{ Coefficient of }{x}^2}}=(-1)\cdot {\displaystyle \frac{b}{a}}$

Solution

(b)

(a) Given that, $2x^2-3x+6=0$

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

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

$\therefore$ Sum of the roots $={\displaystyle \frac{-b}{a}}={\displaystyle \frac{-(-3)}{2}}={\displaystyle \frac{3}{2}}$

So, sum of the roots of the quadratic equation $2x^2-3x+6=0$ is not 3 , so it is not the answer.

(b) Given that, $-x^2+3x-3=0$

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

$\therefore$ Sum of the roots $={\displaystyle \frac{-b}{a}}={\displaystyle \frac{-(3)}{-1}}=3$

So, sum of the roots of the quadratic equation $-x^2+3x-3=0$ is 3 , so it is the answer.

(c) Given that, $\sqrt{2}{x}^2-{\displaystyle \frac{3}{\sqrt{2}}}x+1=0$

$\Rightarrow 2x^2-3x+\sqrt{2}=0$

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

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

$\therefore$ Sum of the roots $={\displaystyle \frac{-b}{a}}={\displaystyle \frac{-(-3)}{2}}={\displaystyle \frac{3}{2}}$

So, sum of the roots of the quadratic equation $\sqrt{2}{x}^2-{\displaystyle \frac{3}{\sqrt{2}}}x+1=0$ is not 3 , so it is not the answer.

(d) Given that, $3x^2-3x+3=0$

$\Rightarrow x^2-x+1=0$

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

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

$\therefore$ Sum of the roots $={\displaystyle \frac{-b}{a}}={\displaystyle \frac{-(-1)}{1}}=1$

So, sum of the roots of the quadratic equation $3x^2-3x+3=0$ is not 3 , so it is not the answer.

Thinking Process

If any quadratic equation i.e., $a{x}^2+bx+c=0,a\ne 0$ has two equal real roots, then its discriminant should be equal to zero. i.e., $D=b^2-4ac=0$

Solution

(d) Given equation is $2x^2-kx+k=0$

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

$a=2,b=-k\text{ and }c=k$

For equal roots, the discriminant must be zero.

i.e.,

$\begin{array}{rl}D=b^2-4ac& =0\\ (-k{)}^2-4(2)k& =0\end{array}$

$\begin{array}{ccc}\Rightarrow & (-k{)}^2-4(2)k& =0\\ \Rightarrow & k^2-8k& =0\\ \Rightarrow & k(k-8)& =0\\ \therefore & k& =0,8\end{array}$

Hence, the required values of $k$ are 0 and 8 .

Solution

(b) Given equation is $9x^2+{\displaystyle \frac{3}{4}}x-\sqrt{2}=0$.

$(3x{)}^2+{\displaystyle \frac{1}{4}}(3x)-\sqrt{2}=0$

On putting $3x=y$, we have $y^2+{\displaystyle \frac{1}{4}}y-\sqrt{2}=0$

$\begin{array}{rl}& y^2+{\displaystyle \frac{1}{4}}y+({\displaystyle \frac{1}{8}}{)}^2-({\displaystyle \frac{1}{8}}{)}^2-\sqrt{2}=0\\ & (y+{\displaystyle \frac{1}{8}}{)}^2={\displaystyle \frac{1}{64}}+\sqrt{2}\\ & (y+{\displaystyle \frac{1}{8}}{)}^2={\displaystyle \frac{1+64\cdot \sqrt{2}}{64}}\end{array}$

Thus, ${\displaystyle \frac{1}{64}}$ must be added and subtracted to solve the given equation.

Thinking Process

If a quadratic equation is in the from of $a{x}^2+bx+c=0,a\ne 0$, then

(i) If D $=b^2-4ac>0$, then its roots are distinct and real.

(ii) If D $=b^2-4ac=0$, then its roots are real and equal.

(iii) If $D=b^2-4ac<0$, then its roots are not real or imaginary roots.

Any quadratic equation must have only two roots.

Solution

(c) Given equation is $2x^2-\sqrt{5}x+1=0$.

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

$\therefore$ Discriminant, $\begin{array}{rl}a& =2,b=-\sqrt{5}\text{ and }c=1\\ D& =b^2-4ac=(-\sqrt{5}{)}^2-4\times (2)\times (1)=5-8\\ & =-3<0\end{array}$

Since, discriminant is negative, therefore quadratic equation $2x^2-\sqrt{5}x+1=0$ has no real roots i.e., imaginary roots.

Solution

(b) The given equation is $x^2+x-5=0$

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

$a=1,b=1\text{ and }c=-5$

The discriminant of $x^2+x-5=0$ is

$$\begin{array}{rl}D& =b^2-4ac=(1{)}^2-4(1)(-5)\\ & =1+20=21\end{array}$$

$\Rightarrow b^2-4ac>0$

So, $x^2+x-5=0$ has two distinct real roots.

(a) Given equation is, $2x^2-3\sqrt{2}x+9/4=0$.

On comparing with $a{x}^2+bx+c=0$

$a=2,b=-3\sqrt{2}\text{ and }c=9/4$

Now, $D=b^2-4ac=(-3\sqrt{2}{)}^2-4(2)(9/4)=18-18=0$

Thus, the equation has real and equal roots.

(c) Given equation is $x^2+3x+2\sqrt{2}=0$

On comparing with $a{x}^2+bx+c=0$

$a=1,b=3\text{ and }c=2\sqrt{2}$

Now, $D=b^2-4ac=(3{)}^2-4(1)(2\sqrt{2})=9-8\sqrt{2}<0$

$\therefore$ Roots of the equation are not real.

(d) Given equation is, $5x^2-3x+1=0$

On comparing with $a{x}^2+bx+c=0$

$a=5,b=-3,c=1$

Now, $D=b^2-4ac=(-3{)}^2-4(5)(1)=9-20<0$

Hence, roots of the equation are not real.

Solution

(a)

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

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

$$a=1,b=-4\text{ and }c=3\sqrt{2}$$

The discriminant of $x^2-4x+3\sqrt{2}=0$ is

$$\begin{array}{rl}D& =b^2-4ac\\ & =(-4{)}^2-4(1)(3\sqrt{2})=16-12\sqrt{2}=16-12\times (1.41)\\ & =16-16.92=-0.92\end{array}$$

$\Rightarrow b^2-4ac<0$

(b) The given equation is $x^2+4x-3\sqrt{2}=0$

On comparing the equation with $a{x}^2+bx+c=0$, we get

Then,

$$\begin{array}{rl}a& =1,b=4\text{ and }c=-3\sqrt{2}\\ D& =b^2-4ac=(-4{)}^2-4(1)(-3\sqrt{2})\\ & =16+12\sqrt{2}>0\end{array}$$

Hence, the equation has real roots.

(c) Given equation is $x^2-4x-3\sqrt{2}=0$

On comparing the equation with $a{x}^2+bx+c=0$, we get

Then,

$$\begin{array}{rl}a& =1,b=-4\text{ and }c=-3\sqrt{2}\\ D& =b^2-4ac=(-4{)}^2-4(1)(-3\sqrt{2})\\ & =16+12\sqrt{2}>0\end{array}$$

Hence, the equation has real roots.

(d) Given equation is $3x^2+4\sqrt{3}x+4=0$.

On comparing the equation with $a{x}^2+bx+c=0$, we get

Then,

$$\begin{array}{rl}& a=3,b=4\sqrt{3}\text{ and }c=4\\ & D=b^2-4ac=(4\sqrt{3}{)}^2-4(3)(4)=48-48=0\end{array}$$

Hence, the equation has real roots.

Hence, $x^2-4x+3\sqrt{2}=0$ has no real roots.

Solution

(c) Given equation is $(x^2+1{)}^2-x^2=0$

$\begin{array}{rl}& \Rightarrow x^4+1+2x^2-x^2=0[\because (a+b{)}^2=a^2+b^2+2ab]\\ & \Rightarrow x^4+x^2+1=0\\ & \text{ Let }{x}^2=y\\ & \therefore (x^2{)}^2+x^2+1=0\\ & y^2+y+1=0\\ & a=1,b=1\text{ and }c=1\\ & \text{ Discriminant, }D=b^2-4ac\\ & =(1{)}^2-4(1)(1)\\ & =1-4=-3\end{array}$

Thinking Process

If $a{x}^2+bx+c=0a\ne 0$ be any quadratic equation and $D=b^2-4ac$ be its discriminant, then

(i) $D>0$ i.e., $b^2-4ac>0\Rightarrow$ Roots are real and distinct.

(ii) $D=0$ i.e., $b^2-4ac=0\Rightarrow$ Roots are real and equal.

(iii) $D<0$ i.e., $b^2-4ac<0\Rightarrow$ No reals roots i.e., imaginary roots.

Solution

(i) Given equation is $x^2-3x+4=0$.

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

$\therefore$ Discriminant, $\begin{array}{rl}a& =1,b=-3\text{ and }c=4\\ D& =b^2-4ac=(-3{)}^2-4(1)(4)\\ & =9-16=-7<0\text{ i.e., }D<0\end{array}$

Hence, the equation $x^2-3x+4=0$ has no real roots.

(ii) Given equation is, $2x^2+x-1=0$

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

$\therefore \text{ Discriminant, }$ $\begin{array}{rl}a& =2,b=1\text{ and }c=-1\\ D& =b^2-4ac=(1{)}^2-4(2)(-1)\\ & =1+8=9>0\text{ i.e., }D<0\end{array}$

Hence, the equation $2x^2+x-1=0$ has two distinct real roots

(iii) Given equation is $2x^2-6x+{\displaystyle \frac{9}{2}}=0$.

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

$\therefore$ Discriminant, $\begin{array}{rl}a& =2,b=-6\text{ and }c={\displaystyle \frac{9}{2}}\\ D& =b^2-4ac\\ & =(-6{)}^2-4(2){\displaystyle \frac{9}{2}}=36-36=0\text{ i.e., }D=0\end{array}$

Hence, the equation $2x^2-6x+{\displaystyle \frac{9}{2}}=0$ has equal and real roots. (iv) Given equation is $3x^2-4x+1=0$.

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

$\begin{array}{rl}& \therefore \text{ Discriminant, }\\ & \begin{array}{rl}a& =3,b=-4\text{ and }c=1\\ D& =b^2-4ac=(-4{)}^2-4(3)(1)\\ & =16-12=4>0\text{ i.e., }D>0\end{array}\end{array}$

Hence, the equation $3x^2-4x+1=0$ has two distinct real roots.

(v) Given equation is $(x+4{)}^2-8x=0$.

$\begin{array}{ccc}\Rightarrow & x^2+16+8x-8x=0& [\because (a+b{)}^2=a^2+2ab+b^2]\\ \Rightarrow & x^2+16=0\\ \Rightarrow & x^2+0\cdot x+16=0\end{array}$

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

$$a=1,b=0\text{ and }c=16$$

$\therefore \text{ Discriminant, }D=b^2-4ac=(0{)}^2-4(1)(16)=-64<0\text{ i.e., }D<0$

Hence, the equation $(x+4{)}^2-8x=0$ has imaginary roots, i.e., no real roots.

(vi) Given equation is

$\begin{array}{rl}& \text{ Given equation is }(x-\sqrt{2}{)}^2-\sqrt{2}(x+1)=0.\\ & \Rightarrow x^2+(\sqrt{2}{)}^2-2x\sqrt{2}-\sqrt{2}x-\sqrt{2}=0[\because (a-b{)}^2=a^2-2ab+b^2]\end{array}$

$\begin{array}{cc}\Rightarrow & x^2+2-2\sqrt{2}x-\sqrt{2}x-\sqrt{2}=0\\ \Rightarrow & x^2-3\sqrt{2}x+(2-\sqrt{2})=0\end{array}$

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

$\therefore$ Discriminant, $D=b^2-4ac$

$$a=1,b=-3\sqrt{2}\text{ and }c=2-\sqrt{2}$$

$$\begin{array}{rl}& =(-3\sqrt{2}{)}^2-4(1)(2-\sqrt{2})=9\times 2-8+4\sqrt{2}\\ & =18-8+4\sqrt{2}=10+4\sqrt{2}>0\text{ i.e., }D>0\end{array}$$

Hence, the equation $(x-\sqrt{2}{)}^2-\sqrt{2}(x+1)=0$ has two distinct real roots.

(vii) Given, equation is $\sqrt{2}{x}^2-{\displaystyle \frac{3}{\sqrt{2}}}x+{\displaystyle \frac{1}{\sqrt{2}}}=0$.

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

$\therefore$ Discriminant, $$a=\sqrt{2},b=-{\displaystyle \frac{3}{\sqrt{2}}}\text{ and }c={\displaystyle \frac{1}{\sqrt{2}}}$$

$$D=b^2-4ac$$

$$\begin{array}{rl}& =-{{\displaystyle \frac{3}{\sqrt{2}}}}^2-4\sqrt{2}{\displaystyle \frac{1}{\sqrt{2}}}\\ & ={\displaystyle \frac{9}{2}}-4={\displaystyle \frac{9-8}{2}}={\displaystyle \frac{1}{2}}>0\text{ i.e., }D>0\end{array}$$

Hence, the equation $\sqrt{2}{x}^2-{\displaystyle \frac{3}{\sqrt{2}}}x+{\displaystyle \frac{1}{\sqrt{2}}}=0$ has two distinct real roots.

(viii) Given equation is $x(1-x)-2=0$.

$\begin{array}{ccc}\Rightarrow & & x-x^2-2=0\\ \Rightarrow & & x^2-x+2=0\end{array}$

On comparing with $a{x}^2+bx+c=0$, we get $a=1,b=-1$ and $c=2$

$\therefore$ Discriminant, $D=b^2-4ac$

$=(-1{)}^2-4(1)(2)=1-8=-7<0\text{ i.e., }D<0$

Hence, the equation $x(1-x)-2=0$ has imaginary roots i.e., no real roots.

(ix) Given equation is

$\begin{array}{cccc}& & (x-1)(x+2)+2& =0\\ \Rightarrow & & x^2+x-2+2& =0\\ \Rightarrow & & x^2+x+0& =0\end{array}$

On comparing the equation with $a{x}^2+bx+c-0$, We have

$a=1,b=1\text{ and }c=0$

$\therefore$ Discriminant, $D=b^2-4ac$

$$=1-4(1)(0)=1>0\text{ i.e., }D>0$$

Hence, equation has two distinct real roots.

(x) Given equation is

$\begin{array}{ccc}& & (x+1)(x-2)+x=0\\ \Rightarrow & & x^2+x-2x-2+x=0\\ \Rightarrow & & x^2-2=0\\ \Rightarrow & & x^2+0\cdot x-2=0\end{array}$

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

$$a=1,b=0\text{ and }c=-2$$

$\therefore$ Discriminant, $D=b^2-4ac=(0{)}^2-4(1)(-2)=0+8=8>0$

Hence, the equation $(x+1)(x-2)+x=0$ has two distinct real roots.

Solution

(i) False, since a quadratic equation has two and only two roots.

(ii) False, for example $x^2+4=0$ has no real root.

(iii) False, since a quadratic equation has two and only two roots.

(iv) True, because every quadratic polynomial has atmost two roots.

(v) True, since in this case discriminant is always positive, so it has always real roots, i.e., ac $<0$ and so, $b^2-4ac>0$.

(vi) True, since in this case discriminant is always negative, so it has no real roots i.e., if $b=0$, then $b^2-4ac\Rightarrow -4ac<0$ and $ac>0$.

Solution

No, consider the quadratic equation $2x^2+x-6=0$ with integral coefficient. The roots of the given quadratic equation are -2 and ${\displaystyle \frac{3}{2}}$ which are not integers.

Solution

Yes, consider the quadratic equation $2x^2+x-4=0$ with rational coefficient. The roots of the given quadratic equation are ${\displaystyle \frac{-1+\sqrt{33}}{4}}$ and ${\displaystyle \frac{-1-\sqrt{33}}{4}}$ are irrational.

Solution

Yes, consider the quadratic equation with all distinct irrationals coefficients i.e., $\sqrt{3}{x}^2-7\sqrt{3}x+12\sqrt{3}=0$. The roots of this quadratic equation are 3 and 4 , which are rationals.

Solution

No, since 0.2 does not satisfy the quadratic equation i.e., $(0.2{)}^2-0.4=0.04-0.4\ne 0$.

Solution

Given that, $b=0$ and $c<0$ and quadratic equation,

$$\begin{array}{}\text{(i)}& x^2+bx+c=0\end{array}$$

Put $b=0$ in Eq. (i), we get

$\begin{array}{ccc}{x}^2+0+c& =0& \\ \Rightarrow & x^2& =-c& \text{ here }c>0\\ \therefore & x& =\pm \sqrt{-c}& \therefore -c>0\end{array}$

Thinking Process

Compare the given quadratic equation with $a{x}^2+bx+c=0$ and get the values of $a,b$ and $c$. Now, use the quadratic formula for finding the roots of quadratic equation.

i.e., $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

Solution

(i) Given equation is $2x^2-3x-5=0$.

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

By quadratic formula,

$\begin{array}{rl}a& =2,b=-3\text{ and }c=-5\\ x& ={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & ={\displaystyle \frac{-(-3)\pm \sqrt{(-3{)}^2-4(2)(-5)}}{2(2)}}={\displaystyle \frac{3\pm \sqrt{9+40}}{4}}\\ & ={\displaystyle \frac{3\pm \sqrt{49}}{4}}={\displaystyle \frac{3\pm 7}{4}}={\displaystyle \frac{10}{4}},{\displaystyle \frac{-4}{4}}={\displaystyle \frac{5}{2}},-1\end{array}$

So, ${\displaystyle \frac{5}{2}}$ and -1 are the roots of the given equation.

(ii) Given equation is $5x^2+13x+8=0$.

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

$a=5,b=13\text{ and }c=8$

By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(13)\pm \sqrt{(13{)}^2-4(5)(8)}}{2(5)}}\\ & ={\displaystyle \frac{-13\pm \sqrt{169-160}}{10}}={\displaystyle \frac{-13\pm \sqrt{9}}{10}}\\ & ={\displaystyle \frac{-13\pm 3}{10}}=-{\displaystyle \frac{10}{10}},-{\displaystyle \frac{16}{10}}=-1,-{\displaystyle \frac{8}{5}}\end{array}$

(iii) Given equation is $-3x^2+5x+12=0$.

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

$a=-3,b=5\text{ and }c=12$

By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(5)\pm \sqrt{(5{)}^2-4(-3)(12)}}{2(-3)}}\\ & ={\displaystyle \frac{-5\pm \sqrt{25+144}}{-6}}={\displaystyle \frac{-5\pm \sqrt{169}}{-6}}\\ & ={\displaystyle \frac{-5\pm 13}{-6}}={\displaystyle \frac{8}{-6}},{\displaystyle \frac{-18}{-6}}=-{\displaystyle \frac{4}{3}},3\end{array}$

So, $-{\displaystyle \frac{4}{3}}$ and 3 are two roots of the given equation.

(iv) Given equation is $-x^2+7x-10=0$.

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

$a=-1,b=7\text{ and }c=-10$

By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(7)\pm \sqrt{(7{)}^2-4(-1)(-10)}}{2(-1)}}={\displaystyle \frac{-7\pm \sqrt{49-40}}{-2}}\\ & ={\displaystyle \frac{-7\pm \sqrt{9}}{-2}}={\displaystyle \frac{-7\pm 3}{-2}}={\displaystyle \frac{-4}{-2}},{\displaystyle \frac{-10}{-2}}=2,5\end{array}$

So, 2 and 5 are two roots of the given equation.

(v) Given equation is $x^2+2\sqrt{2}x-6=0$.

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

$a=1,b=2\sqrt{2}\text{ and }c=-6$

By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(2\sqrt{2})\pm \sqrt{(2\sqrt{2}{)}^2-4(1)(-6)}}{2(1)}}={\displaystyle \frac{-2\sqrt{2}\pm \sqrt{8+24}}{2}}\\ & ={\displaystyle \frac{-2\sqrt{2}\pm \sqrt{32}}{2}}={\displaystyle \frac{-2\sqrt{2}\pm 4\sqrt{2}}{2}}\\ & ={\displaystyle \frac{-2\sqrt{2}+4\sqrt{2}}{2}},{\displaystyle \frac{-2\sqrt{2}-4\sqrt{2}}{2}}=\sqrt{2},-3\sqrt{2}\end{array}$

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

(vi) Given equation is $x^2-3\sqrt{5}x+10=0$.

On comparing with $a{x}^2+bx+c=0$, we have

$a=1,b=-3\sqrt{5}\text{ and }c=10$

By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(-3\sqrt{5})\pm \sqrt{(-3\sqrt{5}{)}^2-4(1)(10)}}{2(1)}}\\ & ={\displaystyle \frac{3\sqrt{5}\pm \sqrt{45-40}}{2}}={\displaystyle \frac{3\sqrt{5}\pm \sqrt{5}}{2}}\\ & ={\displaystyle \frac{3\sqrt{5}+\sqrt{5}}{2}},{\displaystyle \frac{3\sqrt{5}-\sqrt{5}}{2}}=2\sqrt{5},\sqrt{5}\end{array}$

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

(vii) Given equation is ${\displaystyle \frac{1}{2}}{x}^2-\sqrt{11}x+1=0$.

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

$a={\displaystyle \frac{1}{2}},b=-\sqrt{11}\text{ and }c=1$

$\therefore$ By quadratic formula, $x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

$\begin{array}{rl}& ={\displaystyle \frac{-(-\sqrt{11})\pm \sqrt{(-\sqrt{11}{)}^2-4\times {\displaystyle \frac{1}{2}}\times 1}}{2{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{\sqrt{11}\pm \sqrt{11-2}}{1}}=\sqrt{11}\pm \sqrt{9}\\ & =\sqrt{11}\pm 3=3+\sqrt{11},\sqrt{11}-3\end{array}$