Question

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

Answer 1

$$\begin{gathered} ab{x}^2+(b^2-ac)x-bc=0 \\ \text{Here,}a=ab,b=b^2-ac\mathrm{and}c=-bc \\ \text{D =}\sqrt{b^2-4ac} \\ \text{=}\sqrt{{(b^2-ac)}^2-4\times ab\times (-bc)} \\ \text{=}\sqrt{b^4-2b^{2}ac+a^2{c}^2+4b^{2}ac} \\ \text{=}\sqrt{b^4+2b^{2}ac+a^2{c}^2} \\ \text{=}\sqrt{{(b^2+ac)}^2} \\ \text{=}(b^2+ac) \\ \therefore \text{The roots of the equation are:} \\ \frac{-b+\text{D}}{2a},\frac{-b-\text{D}}{2a} \\ \mathrm{or},\frac{-(b^2-ac)+(b^2+ac)}{2ab},\frac{-(b^2-ac)-(b^2+ac)}{2ab} \\ \mathrm{or},\frac{-b^2+ac+b^2+ac}{2ab},\frac{-b^2+ac-b^2-ac}{2ab} \\ \mathrm{or},\frac{2ac}{2ab},\frac{-2b^2}{2ab} \\ \mathrm{or},\frac{c}{b},\frac{-b}{a} \end{gathered}$$