Question

Solve the following quadratic equation by factorization :
$ab{x}^2+(b^2-ac)x-bc=0$

• A. $\frac{b}{a},\frac{b}{c}$
• B. $\frac{-b}{a},\frac{c}{b}$
• C. $\frac{-b}{a},\frac{b}{c}$
• D. $\frac{-b}{a},\frac{-c}{b}$

Answer 1

The correct option is B $\frac{-b}{a},\frac{c}{b}$

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

$\Rightarrow$ $ab{x}^2+b^{2}x-acx-bc=0$

$\Rightarrow$ $bx(ax+b)-c(ax+b)=0$

$\Rightarrow$ $(ax+b)(bx-c)=0$

$\Rightarrow$ $ax+b=0$ and $bx-c=0$

$\therefore$ $x=\frac{-b}{a}$ and $x=\frac{c}{b}$