Question

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

Answer 1

We have,

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

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

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

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

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

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

$\Rightarrow 8x^2-(16+1)x+2=0$

$\Rightarrow 8x^2-16x-1x+2=0$

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

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

$\Rightarrow x-2=08x-1=0$

$\Rightarrow x=2x=\frac{1}{8}$

Hence, this is the answer.