Question

The equation $12x^3+8x^2-x-1=0$ has

• A. $3$ real and distinct roots
• B. $3$ real roots of which two are equal
• C. $2$ positive and $1$ negative roots
• D. $2$ negative and $1$ positive roots

Answer 1

Answer: (B), (D)

$2$ negative and $1$ positive roots

Let $f\left(x\right)=12x^3+8x^2-x-1$
$f^{\prime }\left(x\right)=36x^2+16x-1,f^{\prime \prime }\left(x\right)=72x+16$
$f^{\prime }\left(x\right)=0\Rightarrow x=\frac{1}{18},x=-\frac{1}{2}$
$f(-\frac{1}{2})=0$ and $f^{\prime \prime }(-\frac{1}{2})\ne 0$
$\therefore -\frac{1}{2}$ is a repeated root of $f\left(x\right)=0$
Let $\alpha$ be the third root.
We know, sum of roots $=-\frac{1}{2}-\frac{1}{2}+\alpha =-\frac{8}{12}$
$\Rightarrow \alpha =\frac{1}{3}$
So, the roots of $f\left(x\right)=0$ are $-\frac{1}{2},-\frac{1}{2},\frac{1}{3}$
Clearly, it has $2$ negative and $1$ positive roots.