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

Given:

Given quadratic equation is $abx^2+(b^2-ac)x-bc=0$.

To do:

We have to solve the given quadratic equation.

Solution:

$abx^2+(b^2-ac)x-bc=0$

To factorise $abx^2+(b^2-ac)x-bc=0$, we have to find two numbers $m$ and $n$ such that $m+n=b^2-ac$ and $mn=ab(-bc)=-ab^2c$.

If $m=b^2$ and $n=-ac$, $m+n=b^2-ac$ and $mn=b^2\times(-ac)=-ab^2c$.

$abx^2+(b^2-ac)x-bc=0$

$abx^2+b^2x-acx-bc=0$

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

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

$ax+b=0$ or $bx-c=0$

$ax=-b$ or $bx=c$

$x=-\frac{b}{a}$ or $x=\frac{c}{b}$

The values of $x$ are  $-\frac{b}{a}$ and $\frac{c}{b}$.

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Updated on: 10-Oct-2022

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