Question
Solve the following quadratic equations by factorization:
$a b x^2+\left(b^2-a c\right) x-b c=0$
$a b x^2+\left(b^2-a c\right) x-b c=0$
✓
Answer
We have,
$a b x^2+\left(b^2-a c\right) x-b c=0$
${\left[a b x-b c=-a b^2 c \Rightarrow-a b^2 c=b^2 \times-a c \text { and } b^2-a c=b^2+(-a c)\right]}$
$\Rightarrow a b x^2+b^2 x-a c x-b c=0$
$\Rightarrow b x(a x+b)-c(a x+b)=0$
$\Rightarrow(a x+b)(b x-c)=0$
$\Rightarrow a x+b=0 \text { or } b x-c=0$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
$\therefore\text{x}=-\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{\text{c}}{\text{b}}$ are the two roots the given equation.
$a b x^2+\left(b^2-a c\right) x-b c=0$
${\left[a b x-b c=-a b^2 c \Rightarrow-a b^2 c=b^2 \times-a c \text { and } b^2-a c=b^2+(-a c)\right]}$
$\Rightarrow a b x^2+b^2 x-a c x-b c=0$
$\Rightarrow b x(a x+b)-c(a x+b)=0$
$\Rightarrow(a x+b)(b x-c)=0$
$\Rightarrow a x+b=0 \text { or } b x-c=0$
$\Rightarrow\text{x}=-\frac{\text{b}}{\text{a}}$ or $\text{x}=\frac{\text{c}}{\text{b}}$
$\therefore\text{x}=-\frac{\text{b}}{\text{a}}$ and $\text{x}=\frac{\text{c}}{\text{b}}$ are the two roots the given equation.
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