Question

Solve the following equation:
$\frac{(2x+1{)}^2+(2x-1{)}^2}{(2x+1{)}^2-(2x-1{)}^2}=\frac{17}{8}$

Answer 1

$\frac{{(2x+1)}^2+{(2x-1)}^2}{{(2x+1)}^2-{(2x-1)}^2}=\frac{17}{8}$

(using the formulae ${(a+b)}^2=a^2+b^2+2aband{a}^2-b^2=(a+b)(a-b)$

$\Rightarrow \frac{4x^2+4x+1+4x^2-4x+1}{(2x+1+2x-1)(2x+1-2x+1)}=\frac{17}{8}$

$\Rightarrow \frac{8x^2+2}{\left(4x\right)\left(2\right)}=\frac{17}{8}$

$\Rightarrow \frac{8x^2+2}{8x}=\frac{17}{8}$

Taking $\frac{1}{8}$ common on both the sides,

$\frac{1}{8}\left(\frac{8x^2+2}{x}\right)=\frac{1}{8}\left(17\right)$

$\Rightarrow \frac{8x^2+2}{x}=17$

$\Rightarrow 8x^2+2=17x$

$\Rightarrow 8x^2-17x+2=0$

Factorizing by splitting the middle term,

$\Rightarrow 8x^2-16x-x+2=0$

$\Rightarrow 8x(x-2)-1(x-2)=0$

$\Rightarrow (x-2)(8x-1)=0$

$\therefore x=2,\frac{1}{8}$