Solve the following quadratic equations by factorization:

Solution:

We have been given

$a b x^{2}+\left(b^{2}-a c\right) x-b c=0$

$a b x^{2}+b^{2} x-a c x-b c=0$

$b x(a x+b)+-c(a x+b)=0$

 

$(a x+b)(b x-c)=0$

Therefore,

$a x+b=0$

$a x=-b$

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

or,

$b x-c=0$

$b x=c$

$x=\frac{c}{b}$

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