Question

Given $\frac{m}{n}=\frac{x^2+2y}{x^2-2y}$; make 'x' as the subject of the formula. Then find the value of x, if y = 2, m =5 and n =4.

Answer 1

Given, $\frac{m}{n}-\frac{x^2+2y}{x^2-2y}$
$=>m{x}^2-2my=n{x}^2+2ny$
$=>x^2(m-n)=2y(m+n)$
$x^2=\frac{2y(m+n)}{(m-n)}$
$=>x=\sqrt{\frac{2y(m+n)}{(m-n)}}$
When, $y=2,m=5,n=4,$
$x=\sqrt{\frac{2\left(2\right)(5+4)}{(5-4)}}$
$x=\sqrt{36}$
$x=6$