Question

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

Answer 1

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

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

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

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

$\Rightarrow$ $64x^2+16=136x$

$\Rightarrow$ $64x^2-136x+16=0$

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

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

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

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

$\Rightarrow$ $x-2=0$ and $8x-1=0$

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