Thinking Process

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

(b) Given that,

$ \begin{matrix}& -2 x^{2} & =(5-x) 2 x-\dfrac{2}{5} \\ \Rightarrow & &-2 x^{2} & =10 x-2 x^{2}-2+\dfrac{2 x}{5} \\ \Rightarrow & &50 x+2 x-10 & =0 \\ \Rightarrow & &52 x-10 & =0 \end{matrix} $

(c) Given that,

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

which is also not a quadratic equation.

(d)

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

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

$a x^{2}+b x+c=0, a \neq 0$.

Solution

(d)

(a)

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

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

(b)

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

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

(c)

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

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

(d) Given that,

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

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

Thinking Process

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

Solution

(c)

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

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

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

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

$$ \begin{aligned} & (2)^{2}+3(2)-12 \\ & =4+6-12=-2 \neq 0 \end{aligned} $$

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

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

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

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

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

$$ \begin{aligned} & 3(2)^{2}-6(2)-2 \\ & =12-12-2=-2 \neq 0 \end{aligned} $$

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

Thinking Process

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

Solution

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

$ \begin{matrix} \text{ Then, } & & (\dfrac{1}{2})^{2}+k (\dfrac{1}{2})-\dfrac{5}{4} =0 \\ \Rightarrow & & \dfrac{1}{4}+\dfrac{k}{2}-\dfrac{5}{4} =0 \Rightarrow \dfrac{1+2 k-5}{4}=0 \\ \Rightarrow & &2 k-4 =0 \\ \Rightarrow & &2 k =4 \Rightarrow k=2 \end{matrix} $

Thinking Process

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

Solution

(b)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \quad x^{2}-x+1=0$

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

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

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

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

Thinking Process

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

Solution

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

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

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

For equal roots, the discriminant must be zero.

i.e.,

$ \begin{aligned} D=b^{2}-4 a c & =0 \\ (-k)^{2}-4(2) k & =0 \end{aligned} $

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

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

Solution

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

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

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

$ \begin{aligned} & y^{2}+\dfrac{1}{4} y+\Big(\dfrac{1}{8} \Big)^{2}-\Big(\dfrac{1}{8}\Big)^{2}-\sqrt{2}=0 \\ & \Big(y+\dfrac{1}{8}\Big)^{2}=\dfrac{1}{64}+\sqrt{2} \\ & \Big(y+\dfrac{1}{8}\Big)^{2}=\dfrac{1+64 \cdot \sqrt{2}}{64} \end{aligned} $

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

Thinking Process

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

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

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

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

Any quadratic equation must have only two roots.

Solution

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

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

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

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

Solution

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

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

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

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

$$ \begin{aligned} D & =b^{2}-4 a c=(1)^{2}-4(1)(-5) \\ & =1+20=21 \end{aligned} $$

$ \Rightarrow \hspace{20 mm} b^{2}-4 a c>0 $

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

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

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

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

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

Thus, the equation has real and equal roots.

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

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

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

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

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

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

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

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

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

Hence, roots of the equation are not real.

Solution

(a)

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

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

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

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

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

$ \Rightarrow \hspace{20 mm} b^{2}-4 a c<0 $

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

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

Then,

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

Hence, the equation has real roots.

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

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

Then,

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

Hence, the equation has real roots.

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

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

Then,

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

Hence, the equation has real roots.

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

Solution

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

$ \begin{aligned} & \Rightarrow \quad x^{4}+1+2 x^{2}-x^{2}=0 \quad[\because(a+b)^{2}=a^{2}+b^{2}+2 a b] \\ & \Rightarrow \quad x^{4}+x^{2}+1=0 \\ & \text{ Let } \quad x^{2}=y \\ & \therefore \quad(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}-4 a c \\ & =(1)^{2}-4(1)(1) \\ & =1-4=-3 \end{aligned} $

Thinking Process

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

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

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(vi) Given equation is

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

$ \begin{matrix} \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{matrix} $

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

$\therefore$ Discriminant, $D=b^{2}-4 a c$

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

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

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

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

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

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

$$ D=b^{2}-4 a c $$

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

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

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

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

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

$\therefore$ Discriminant, $D=b^{2}-4 a c$

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

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

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

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

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

Hence, equation has two distinct real roots.

(x) Given equation is

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

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

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

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

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, $b^{2}-4 a c>0$.

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

Solution

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

Solution

Yes, consider the quadratic equation $2 x^{2}+x-4=0$ with rational coefficient. The roots of the given quadratic equation are $\dfrac{-1+\sqrt{33}}{4}$ and $\dfrac{-1-\sqrt{33}}{4}$ 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., $(0.2)^{2}-0.4=0.04-0.4 \neq 0$.

Solution

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

$$ \begin{equation*} x^{2}+b x+c=0 \tag{i} \end{equation*} $$

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

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

Thinking Process

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

i.e., $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Solution

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

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

By quadratic formula,

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

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

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

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

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

By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

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

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

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

By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

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

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

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

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

By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

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 $a x^{2}+b x+c=0$, we get

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

By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

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 $a x^{2}+b x+c=0$, we have

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

By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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

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

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

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

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

$\therefore$ By quadratic formula, $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$ \begin{aligned} & =\dfrac{-(-\sqrt{11}) \pm \sqrt{(-\sqrt{11})^{2}-4 \times \dfrac{1}{2} \times 1}}{2 \dfrac{1}{2}} \\ & =\dfrac{\sqrt{11} \pm \sqrt{11-2}}{1}=\sqrt{11} \pm \sqrt{9} \\ & =\sqrt{11} \pm 3=3+\sqrt{11}, \sqrt{11}-3 \end{aligned} $

So, $3+\sqrt{11}$ and $\sqrt{11}-3$ are the roots of the given equation.

Thinking Process

If any coefficient of the quadratic equation of the form $a x^{2}+b x+c=0$, is in fractional form, then make all the coefficients in integral form. Then, use the factorisation method i.e., by splitting the middle term, we get the required roots of the given quadratic equation.

Solution

(i) Given equation is $2 x^{2}+\dfrac{5}{3} x-2=0$

On multiplying by 3 on both sides, we get

$ \begin{matrix} \Rightarrow & 6 x^{2}+5 x-6 & =0 \\ \Rightarrow & 6 x^{2}+(9 x-4 x)-6 & =0 \\ \Rightarrow & 6 x^{2}+9 x-4 x-6 & =0 \\ \Rightarrow & 3 x(2 x+3)-2(2 x+3) & =0 \\ \text{ Now, } & (2 x+3)(3 x-2) & =0 \\ \Rightarrow & 2 x+3 & =0 \\ \text{ and } & x & =-\dfrac{3}{2} \\ \Rightarrow & 3 x-2 & =0 \\ & x & =\dfrac{2}{3} \end{matrix} $

Hence, the roots of the equation $6 x^{2}+5 x-6=0$ are $\dfrac{-3}{2}$ and $\dfrac{2}{3}$. (ii) Given equation is $\dfrac{2}{5} x^{2}-x-\dfrac{3}{5}=0$.

On multiplying by 5 on both sides, we get

$ \begin{matrix} \Rightarrow & 2 x^{2}-5 x-3 & =0 \\ \Rightarrow & 2 x^{2}-(6 x-x)-3 & =0 \\ \Rightarrow & 2 x^{2}-6 x+x-3 & =0 \\ \Rightarrow & 2 x(x-3)+1(x-3) & =0 \\ \text{ Now, } & (x-3)(2 x+1) & =0 \\ \text{ and } & x-3=0 \Rightarrow x & =3 \\ \Rightarrow & 2 x+1 & =0 \\ & x & =-\dfrac{1}{2} \end{matrix} $

$ 2 x^{2}-(6 x-x)-3=0 \quad \text{ [by splitting the middle term] } $

Hence, the roots of the equation $2 x^{2}-5 x-3=0$ are $-\dfrac{1}{2}$ and 3 .

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

$ \begin{matrix}& 3 \sqrt{2} x^{2}-(6 x-x)-\sqrt{2} =0 & \text{ [by splitting the middle term] } \\ & 3 \sqrt{2} x^{2}-6 x+x-\sqrt{2} =0 \\ & 3 \sqrt{2} x^{2}-3 \sqrt{2} \cdot \sqrt{2} x+x-\sqrt{2} =0 \\ \Rightarrow & 3 \sqrt{2} x(x-\sqrt{2)}+1(x-\sqrt{2}) =0 \\ \Rightarrow & (x-\sqrt{2})(3 \sqrt{2} x+1) =0 \\ \text{ Now, } & x-\sqrt{2}=0 \Rightarrow x =\sqrt{2} \\ \text{ and } & 3 \sqrt{2} x+1 =0 \\ \Rightarrow & x =-\dfrac{1}{3 \sqrt{2}}=\dfrac{-\sqrt{2}}{6} \end{matrix} $

Hence, the roots of the equation $3 \sqrt{2} x^{2}-5 x-\sqrt{2}=0$ are $-\dfrac{\sqrt{2}}{6}$ and $\sqrt{2}$.

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

$ \begin{aligned} & 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-2 \sqrt{5} \cdot \sqrt{5}=0 \quad \text{ [by splitting the middle term] } \\ \Rightarrow & 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-10=0 \\ & 3 x^{2}+6 \sqrt{5} x-\sqrt{5} x-2 \sqrt{5} \cdot \sqrt{5}=0 \\ \Rightarrow & 3 x(x+2 \sqrt{5})-\sqrt{5}(x+2 \sqrt{5})=0 \\ \Rightarrow & x+2 \sqrt{5})(3 x-\sqrt{5})=0 \\ \text{ Now, } & x+2 \sqrt{5}=0 \\ \Rightarrow & x=-2 \sqrt{5} \text{ and } 3 x-\sqrt{5}=0 \\ \Rightarrow & x=\dfrac{\sqrt{5}}{3} \end{aligned} $

Hence, the roots of the equation $3 x^{2}+5 \sqrt{5} x-10=0$ are $-2 \sqrt{5}$ and $\dfrac{\sqrt{5}}{3}$.

(v) Given equation is $21 x^{2}-2 x+\dfrac{1}{21}=0$.

On multiplying by 21 on both sides, we get

$$ 441 x^{2}-42 x+1=0 $$

$$ 441 x^{2}-(21 x+21 x)+1=0 \quad \text{ [by splitting the middle term] } $$

$ \begin{matrix} \Rightarrow & 441 x^{2}-21 x-21 x+1=0 \\ \Rightarrow & 21 x(21 x-1)-1(21 x-1)=0 \\ \Rightarrow & (21 x-1)(21 x-1)=0 \\ \text{ Now, } & 21 x-1=0 \Rightarrow \quad x=\dfrac{1}{21} \text{ and } 21 x-1=0 \\ \therefore & x=\dfrac{1}{21} \end{matrix} $

Hence, the roots of the equation $441 x^{2}-42 x+1=0$ are $\dfrac{1}{21}$ and $\dfrac{1}{21}$.

Thinking Process

(i) Firstly we will check the quadratic equation has real roots or not for this, is discriminant, $D=b^{2}-4 a c>0 \Rightarrow$ roots are real.

(ii) If roots are real, we may factorise the equation or use the quadratic formula to obtain the roots of the equation.

Solution

(i) Given equation is $8 x^{2}+2 x-3=0$.

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

$ a=8, b=2 \text{ and } c=-3 $

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c$

$$ \begin{aligned} & =(2)^{2}-4(8)(-3) \\ & =4+96=100>0 \end{aligned} $$

Therefore, the equation $8 x^{2}+2 x-3=0$ has two distinct real roots because we know that, if the equation $a x^{2}+b x+c=0$ has discriminant greater than zero, then if has two distinct real roots

Roots

$$ \begin{aligned} x & =\dfrac{-b \pm \sqrt{D}}{2 a}=\dfrac{-2 \pm \sqrt{100}}{16}=\dfrac{-2 \pm 10}{16} \\ & =\dfrac{-2+10}{16}, \dfrac{-1-10}{16} \\ & =\dfrac{8}{16},-\dfrac{12}{16}=\dfrac{1}{2},-\dfrac{3}{4} \end{aligned} $$

(ii) Given equation is $-2 x^{2}+3 x+2=0$.

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

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

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c$

$$ \begin{aligned} & =(3)^{2}-4(-2)(2) \\ & =9+16=25>0 \end{aligned} $$

Therefore, the equation $-2 x^{2}+3 x+2=0$ has two distinct real roots because we know that if the equation $a x^{2}+b x+c=0$ has its discriminant greater than zero, then it has two distinct real roots.

$$ \text{ Roots, } \begin{aligned} x & =\dfrac{-b \pm \sqrt{D}}{2 a}=\dfrac{-3 \pm \sqrt{25}}{2(-2)} \\ & =\dfrac{-3 \pm 5}{-4}=\dfrac{-3+5}{-4}, \dfrac{-3-5}{-4} \\ & =\dfrac{2}{-4}, \dfrac{-8}{-4}=-\dfrac{1}{2}, 2 \end{aligned} $$

(iii) Given equation is $5 x^{2}-2 x-10=0$.

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

$$ a=5, b=-2 \text{ and } c=-10 $$

$\therefore$ Discriminant, $D=b^{2}-4 a c$

$$ \begin{aligned} & =(-2)^{2}-4(5)(-10) \\ & =4+200=204>0 \end{aligned} $$

Therefore, the equation $5 x^{2}-2 x-10=0$ has two distinct real roots.

$$ \text{ Roots, } \begin{aligned} x & =\dfrac{-b \pm \sqrt{D}}{2 a} \\ & =\dfrac{-(-2) \pm \sqrt{204}}{2 \times 5}=\dfrac{2 \pm 2 \sqrt{51}}{10} \\ & =\dfrac{1 \pm \sqrt{51}}{5}=\dfrac{1+\sqrt{51}}{5}, \dfrac{1-\sqrt{51}}{5} \end{aligned} $$

(iv) Given equation is $\dfrac{1}{2 x-3}+\dfrac{1}{x-5}=1, x \neq \dfrac{3}{2}$, 5

$ \begin{matrix} \Rightarrow & \dfrac{x-5+2 x-3}{(2 x-5)(x-5)}=1 \\ \Rightarrow & \dfrac{3 x-8}{2 x^{2}-5 x-10 x+25}=1 \\ \Rightarrow & \dfrac{3 x-8}{2 x^{2}-15 x+25}=1 \\ \Rightarrow & 3 x-8=2 x^{2}-15 x+25 \\ \Rightarrow & 2 x^{2}-15 x-3 x+25+8=0 \\ \Rightarrow & 2 x^{2}-18 x+33=0 \end{matrix} $

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

$ a=2, b=-18 \text{ and } c=33 $

$\therefore$ Discriminant, $D=b^{2}-4 a c$

$$ \begin{aligned} & =(-18)^{2}-4 \times 2(33) \\ & =324-264=60>0 \end{aligned} $$

Therefore, the equation $2 x^{2}-18 x+33=0$ has two distinct real roots.

$$ \text{ Roots, } \begin{aligned} x & =\dfrac{-b \pm \sqrt{D}}{2 a}=\dfrac{-(-18) \pm \sqrt{60}}{2(2)} \\ & =\dfrac{18 \pm 2 \sqrt{15}}{4}=\dfrac{9 \pm \sqrt{15}}{2} \\ & =\dfrac{9+\sqrt{15}}{2}, \dfrac{9-\sqrt{15}}{2} \end{aligned} $$

(v) Given equation is $x^{2}+5 \sqrt{5} x-70=0$.

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

$$ a=1, b=5 \sqrt{5} \text{ and } c=-70 $$

$\therefore \quad$ Discriminant, $D=b^{2}-4 a c=(5 \sqrt5^{2}-4(1)(-70).$

$$ =125+280=405>0 $$

Therefore, the equation $x^{2}+5 \sqrt{5} x-70=0$ has two distinct real roots.

$$ \text{ Roots, } \begin{aligned} x & =\dfrac{-b \pm \sqrt{D}}{2 a} \\ & =\dfrac{-5 \sqrt{5} \pm \sqrt{405}}{2(1)}=\dfrac{-5 \sqrt{5} \pm 9 \sqrt{5}}{2} \\ & =\dfrac{-5 \sqrt{5}+9 \sqrt{5}}{2}, \dfrac{-5 \sqrt{5}-9 \sqrt{5}}{2} \\ & =\dfrac{4 \sqrt{5}}{2},-\dfrac{14 \sqrt{5}}{2}=2 \sqrt{5},-7 \sqrt{5} \end{aligned} $$

Thinking Process

Firstly, we make a quadratic equation by using the given condition after that we solve the equation by factorisation equation. Finally, method to obtain the desired number.

Solution

Let $n$ be a required natural number.

Square of a natural number diminished by $84=n^{2}-84$

and thrice of 8 more than the natural number $=3(n+8)$

Now, by given condition,

Hence, the required natural number is 12 .

Solution

Let the natural number be $x$.

According to the question,

$ x+12=\dfrac{160}{x} $

On multiplying by $x$ on both sides, we get

$ \begin{matrix} \Rightarrow & x^{2}+12 x-160=0 \\ \Rightarrow & x^{2}+(20 x-8 x)-160=0 \\ \Rightarrow & x^{2}+20 x-8 x-160=0 \\ \Rightarrow & x(x+20)-8(x+20)=0 \\ \Rightarrow & (x+20)(x-8)=0 \end{matrix} $

[by factorisation method]

Now, $x+20=0 \Rightarrow x=-20$ which is not possible because natural number is always greater than zero and $x-8=0 \Rightarrow x=8$.

Hence, the required natural number is 8 .

Solution

Let the original speed of the train $=x km / h$

Then, the increased speed of the train $=(x+5) km / h$ [by given condition] and distance $=360 km$

According to the question,

$ \begin{aligned} & \dfrac{360}{x}-\dfrac{360}{x+5}=\dfrac{4}{5} \\ & \because \text{ time }=\dfrac{\text{ Distance }}{\text{ Speed }} \\ & \text{ and } 48 \min =\dfrac{48}{60} h=\dfrac{4}{5} h \\ & \Rightarrow \quad \dfrac{360(x+5)-360 x}{x(x+5)}=\dfrac{4}{5} \quad \because 48 \min =\dfrac{48}{60} h=\dfrac{4}{5} h \\ & \Rightarrow \quad \dfrac{360 x+1800-360 x}{x^{2}+5 x}=\dfrac{4}{5} \\ & \Rightarrow \quad \dfrac{1800}{x^{2}+5 x}=\dfrac{4}{5} \\ & \Rightarrow \quad x^{2}+5 x=\dfrac{1800 \times 5}{4}=2250 \\ & \Rightarrow \quad x^{2}+5 x-2250=0 \\ & x^{2}+(50 x-45 x)-2250=0 \\ & x^{2}+50 x-45 x-2250=0 \quad \text{ [by factorisation method] } \\ & x(x+50)-45(x+50)=0 \\ & (x+50)(x-45)=0 \end{aligned} $

Solution

Let the actual age of Zeba $=x yr$.

Herage when she was 5 yr younger $=(x-5)$ yr.

Now, by given condition,

Square of her age $=11$ more than five times her actual age

$ \begin{matrix} & & (x-5)^{2} =5 \text{ actual age }+11 \\ \Rightarrow & & (x-5)^{2} =5 x+11 \\ \Rightarrow & & x^{2}+25-10 x =5 x+11 \\ \Rightarrow & & x^{2}-15 x+14 =0 \\ \Rightarrow & & x^{2}-14 x-x+14 =0 \\ \Rightarrow & & x(x-14)-1(x-14) =0 \\ \Rightarrow & & (x-1)(x-14) =0 \\ \Rightarrow & & x =14 \end{matrix} $

[here, $x \neq 1$ because her age is $x-5$. So, $x-5=1-5=-4$ i.e., age cannot be negative] Hence, required Zeba’s age now is $14 yr$.

Solution

Let Nisha’s present age be $x$ yr.

Then, Asha’s present age $=x^{2}+2$

[by given condition]

Now, when Nisha grows to her mother’s present age i.e., after ${(x^{2}+2)-x}$ yr. Then,

Asha’s age also increased by $[(x^{2}+2)-x] y$ r.

Again by given condition,

Age of Asha $=$ One years less than 10 times the present age of Nisha

$ \begin{matrix} \Rightarrow (x^{2}+2)+{(x^{2}+2)-x} =10 x-1 \\ \Rightarrow 2x^{2}-x+4 =10 x-1 \\ \Rightarrow 2x^{2}-11 x+5 =0 \\ \Rightarrow 2x^{2}-10 x-x+5 =0 \\ \Rightarrow x =5 \\ & (x-5)(2 x-1) =0 \\ [\text{ here, } x=\dfrac{1}{2} \text{ cannot be possible, because at } x=\dfrac{1}{2} \text{, Asha’s age is } 2 \dfrac{1}{4} \text{ yr which is not possible }] \end{matrix} $

Hence, required age of Nisha $=5 yr$

and required age of Asha $=x^{2}+2=(5)^{2}+2=25+2=27 yr$

Solution

Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions $50 m \times 40 m$. So, the distance between pond and lawn would be same around the pond. Say $x m$.

Now, length of rectangular lawn $(l_1)=50 m$

and breadth of rectangular lawn $(b_1)=40 m$

$\therefore$ Length of rectangular pond $(l_2)=50-(x+x)=50-2 x$

and breadth of rectangular pond $(b_2)=40-(x+x)=40-2 x$

Also, area of the grass surrounding the pond $=1184 m^{2}$

$\therefore$ Area of rectangular lawn - Area of rectangular pond

$=$ Area of grass surrounding the pond $l_1 \times b_1-l_2 \times b_2=1184 \quad[\because$ area of rectangle $=$ length $\times$ breadth $]$

$\Rightarrow \quad 50 \times 40-(50-2 x)(40-2 x)=1184$

$\Rightarrow \quad 2000-(2000-80 x-100 x+4 x^{2})=1184$

$\Rightarrow \quad 80 x+100 x-4 x^{2}=1184$

$\Rightarrow \quad 4 x^{2}-180 x+1184=0$

$\Rightarrow \quad x^{2}-45 x+296=0$

$\Rightarrow \quad x^{2}-37 x-8 x+296=0 \quad$ [by splitting the middle term]

$\Rightarrow \quad x(x-37)-8(x-37)=0$

$\Rightarrow \quad(x-37)(x-8)=0$

$ \begin{aligned} & \therefore \quad x=8 \\ & \text{ [At } x=37 \text{, length and breadth of pond are }-24 \text{ and }-34 \text{, respectively but length and } \\ & \text{ breadth cannot be negative. So, } x=37 \text{ cannot be possible] } \\ & \therefore \quad \text{ Length of pond }=50-2 x=50-2(8)=50-16=34 m \\ & \text{ and breadth of pond }=40-2 x=40-2(8)=40-16=24 m \end{aligned} $

Solution

We know that, the time between $2 pm$ to $3 pm=1 h=60 min$

Given that, at $t$ min past $2 pm$, the time needed by the min hand of a clock to show $3 pm$ was found to be 3 min less than $\dfrac{t^{2}}{4} \min$ i.e.,

$ \begin{matrix} & t+\dfrac{t^{2}}{4}-3 =60 \\ \Rightarrow & 4 t+t^{2}-12 =240 \\ \Rightarrow & t^{2}+4 t-252 =0 \\ \Rightarrow & t^{2}+18 t-14 t-252 =0 \\ \Rightarrow & t(t+18)-14(t+18) =0 \\ \Rightarrow & (t+18)(t-14) =0 \\ \therefore & t =14 \text{ min } \end{matrix} $

$ \begin{matrix} \Rightarrow & t^{2}+18 t-14 t-252=0 & \text{ [by splitting the middle term] } \\ \Rightarrow & t(t+18)-14(t+18)=0 & \text{ [since, time cannot be negative, so } t \neq-18] \end{matrix} $

Hence, the required value of $t$ is $14 min$.