Question
Solve the following quadratic equations by factorization:
$ax^2 + (4a^2 - 3b)x - 12ab = 0$

Answer

We have,
$a x^2+\left(4 a^2-3 b\right) x-12 a b=0$
${\left[a \times 12 a b=-12 a^2 b^2=4 a^2 x-3 b\right]}$
$\Rightarrow a x^2+4 a^2 x-3 b x+(4 a \times(-3 b))=0$
$\Rightarrow a x(x+4 a)-3 b(x+4 a)=0$
$\Rightarrow(x+4 a)(a x-3 b)=0$
$\Rightarrow(x+4 a)=0 \text { or }(a x-3 b)=0$
$\Rightarrow x=-4 a \text { or } x=\frac{3 b}{a}$
$\therefore x =\frac{3 b}{ a }$ and $x =-4 a$ are the two roots of the given equations.

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