Question

Using quadratic formula, solve the following equation for x :

$ab{x}^2+(b^2-ac)x-bc=0$

Answer 1

$ab{x}^2+(b^2-ac)x-bc=0$

$\therefore x=\frac{-(b^2-ac)\pm \sqrt{(b^2-ac{)}^2-4(ab\left)\right(-bc)}}{2ab}$

$x=\frac{-(b^2-ac)\pm \sqrt{(b^2-ac{)}^2+4a{b}^{2}c}}{2ab}$

$x=\frac{-(b^2-ac)\pm \sqrt{(b^2+ac{)}^2}}{2ab}$

$x=\frac{-(b^2-ac)+(b^2+ac)}{2ab},x=\frac{-(b^2-ac)-(b^2+ac)}{2ab}$

$x=\frac{2ac}{2ab},x=\frac{-2b^2}{2ab}\Rightarrow x=\frac{c}{b},x=\frac{-b}{a}$