Solve the following quadratic equation by factorization:
$ax^2\ +\ (4a^2\ –\ 3b)x\ –\ 12ab\ =\ 0$

Given:

Given quadratic equation is $ax^2\ +\ (4a^2\ –\ 3b)x\ –\ 12ab\ =\ 0$.

To do:

We have to solve the given quadratic equation by factorization. 

Solution:

$ax^2+(4a^2-3b)x-12ab=0$

$ax^2+4a^2x-3bx-12ab=0$

$ax(x+4a)-3b(x+4a)=0$

$(ax-3b)(x+4a)=0$

$ax-3b=0$ or $x+4a=0$

$ax=3b$ or $x=-4a$

$x=\frac{3b}{a}$ or $x=-4a$

The roots of the given quadratic equation are $\frac{3b}{a}$ and $-4a$.  

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Updated on: 10-Oct-2022

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