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. Only Statement $1$ is possible.
• B. Both Statements are possible.
• C. Only Statement $2$ is possible.
• D. Both Statements are "NOT" possible.

Answer 1

The correct option is A Only Statement $1$ is possible.

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.