Question

If the sum of the two roots of the equation $4x^3+16x^2-9x-36=0$ is zero, then the roots are

• A. 1, 2, -2
• B. -3, $\frac{3}{2},-\frac{3}{2}$
• C. -4, $\frac{3}{2},-\frac{3}{2}$
• D. -2, $\frac{2}{3},-\frac{2}{3}$

Answer 1

The correct option is C

-4, $\frac{3}{2},-\frac{3}{2}$

Let roots of $4x^3+16x^2-9x-36=0$ be $\alpha ,-\alpha ,\beta$.

$\text{Sum of roots =}\alpha -\alpha +\beta =\beta =-\frac{16}{4}=-4$
Hence $x=-4$ is a root of the equation. Using long division method, reduced equation is
$4x^2(x+4)-9(x+4)=0$
⇒ $(x+4)(4x^2-9)=0$
⇒ $x=-4,x=\pm \frac{3}{2}$