Question

Using quadratic formula, solve the given equation for $x$
$ab{x}^2+(b^2-ac)x-bc=0$

Answer 1

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

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

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

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

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