Question

Solve the following quadratic equation by factorization, the roots are $\frac{3b}{a}$, $-4a$:
$a{x}^2+(4a^2-3b)x-12ab=0$

• A. True
• B. False

Answer 1

The correct option is A True

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

$\Rightarrow$ $a{x}^2+4a^{2}x-3bx-12ab=0$

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

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

$\Rightarrow$ $ax-3b=0$ and $x+4a=0$

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

$\therefore$ Roots given in question are correct.