Question

Consider the cubic equation $x^3-4x^2+x+6=0$
Statement $1:$ Equation has $2$ Positive real roots, $1$ Negative real root
Statement $2:$ Equation has $1$ Negative real root, $2$ Imaginary roots.

• A. Possibility no. 1
• B. Possibility no. 2
• C. Both of the above

Answer 1

The correct option is A Possibility no. 1

Given that for the cubic equation $x^3-4x^2+x+6=0,$
Possibility no$.1$ is: $2$ Positive real roots, $1$ Negative real root
and Possibility no$.2$ is: $1$ Negative real root, $2$ Imaginary roots.

Let $f\left(x\right)=x^3-4x^2+x+6$
Number of sign changes are $2,$
This means $f\left(x\right)$ can have either $0\text{or}2$ positive roots.
Now, $f\left(2\right)=8-16+2+6=0$
$\Rightarrow 2$ is one of the roots of the $f\left(x\right).$
Thus, there should exist one more positive real root.
Also, $f(-x)=-x^3-4x^2-x+6$ which means just $1$ sign change.
Thus $f\left(x\right)$ should have one sign change.
Thus, Possibility $1$ is correct.