Question

Solve quadratic equation by polarization $\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$.

Answer 1

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7}$

$\Rightarrow \frac{(x-3{)}^2-(x+3{)}^2}{x^2-9}=\frac{48}{7}$

$\Rightarrow \frac{x^2+9-6x-x^2-9-6x}{x^2-9}=\frac{48}{7}$

$\Rightarrow \frac{-12x}{x^2-9}=\frac{48}{7}$

$\Rightarrow \frac{-x}{x^2-9}=\frac{4}{7}$

$\Rightarrow -7x=4x^2-36$

$\Rightarrow 4x^2+7x-36=0$

$\Rightarrow 4x^2+16x-9x-36=0$

$\Rightarrow 4x(x+4)-9(x+4)=0$

$\Rightarrow (x+4)(4x-9)=0$

$\Rightarrow x=-4,\frac{9}{4}$