Find the roots of the following equations :

Solution:

(i) $x-\frac{1}{x}=3 \quad \Rightarrow x^{2}-3 x-1=0$

On comparing this equation with

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

we obtain $\mathrm{a}=1, \mathrm{~b}=-3, \mathrm{c}=-1$

By using quadratic formula, we obtain

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \Rightarrow x=\frac{3 \pm \sqrt{9+4}}{2}$

$\Rightarrow x=\frac{3 \pm \sqrt{13}}{2}$

Therefore, $x=\frac{3+\sqrt{13}}{2}$ or $\frac{3-\sqrt{13}}{2}$

(ii) $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30} ; x \neq-4$ and $x \neq 7$

$\Rightarrow \frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30} \Rightarrow \frac{-11}{\left(x^{2}-3 x-28\right)}=\frac{11}{30}$

$\Rightarrow-\left(x^{2}-3 x-28\right)=30$

$\Rightarrow x^{2}-3 x-28+30=0$

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

$a=1, b=-3, c=2$

$D=(-3)^{2}-4(1)(2)=1 \Rightarrow \sqrt{\mathbf{D}}=\mathbf{1}$

The two roots are given by $\frac{-\mathbf{b} \pm \sqrt{\mathbf{D}}}{\mathbf{2 a}}$, ie, $\frac{\mathbf{3} \pm \mathbf{1}}{\mathbf{2}}$.

Hence, the two roots are 1 and 2 .