Question

Find the roots of the following equations:
(i) $x-\frac{1}{x}=3,x\ne 0$
(ii) $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x\ne -4,7$

Answer 1

i)

Given equation $x-\frac{1}{x}=3$

$\Rightarrow \frac{x^2-1}{x}=3$

$\Rightarrow x^2-1=3x$

$\Rightarrow x^2-3x-1=0$
Here, the middle term$-3x$ can't be expresses as sum of two terms such that the their product is equal to product of extreme terms.

So, we use the formula of $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

$\Rightarrow x=\frac{-(-3)\pm \sqrt{(-3{)}^2-4(1\left)\right(-1)}}{2\left(1\right)}$

$\Rightarrow x=\frac{3\pm \sqrt{9+4}}{2}$

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

ii) Similarly,

For $\frac{1}{(x+4)}-\frac{1}{(x-7)}=\frac{11}{30}$

$\Rightarrow \frac{(x-7-x-4)}{(x+4)(x-7)}=\frac{11}{30}$

$\Rightarrow \frac{-11}{(x+4)(x-7)}=\frac{11}{30}$

$\Rightarrow -30=x^2-3x-28$

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

Here, the middle term $-3x$ can be expressed as sum of $(-2x)$ and $(-x)$ such that their product $(-2x)\times (-x)=2x^2$ is equal to product of extreme terms ($x^2\times 2=2x^2$)

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

$\Rightarrow x(x-2)-1(x-2)=0$

$\Rightarrow (x-2)(x-1)=0$

$\therefore x=1,2$