Question

If ${abx}^2+(b^2-ac)x-bc=0$, then value of $x$ is equal to:

• A. $\frac{-b}{a},\frac{c}{b}$
• B. $a,c$
• C. $a,b$
• D. $0$

Answer 1

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

Give equation is:

${abx}^2+(b^2-ac)x-bc=0$
$\Rightarrow {abx}^2+b^{2}x-acx-bc=0$
$\Rightarrow bx(ax+b)-c(ax+b)=0$
$\Rightarrow (ax+b)(bx-c)=0$
$\Rightarrow ax+b=0orbx-c=0$
$\Rightarrow x=-\frac{b}{a}$ and $x=\frac{c}{b}$
Hence, $x=\frac{c}{b},-\frac{b}{a}$ are required solution.