Question

Solve the following equation:
$\frac{4}{x+\sqrt{x^2+x}}-\frac{1}{x-\sqrt{x^2+x}}=\frac{3}{x}$

Answer 1

$\frac{4}{x+\sqrt{x^2+x}}-\frac{1}{x-\sqrt{x^2+x}}=\frac{3}{x}$

$\frac{4x-4\sqrt{x^2+x}-x-\sqrt{x^2+x}}{x^2-(x^2+x)}=\frac{3}{x}$

$\frac{3x-5\sqrt{x^2+x}}{x}=\frac{3}{x}$

$x\ne 0$

$\therefore 3x-5\sqrt{x^2+x}=-3$

$3(x+1)=5\sqrt{x^2+x}$

Squaring both sides

$9(x^2+2x+1)=25x^2+25x$

$16x^2+7x-9=0$

$16x^2+16x-9x-9=0$

$(16x-9)(x+1)=0$

$x=\frac{9}{16},-1$