Question

If $4x^6+a{x}^3+5x-7=0$ has a maximum of $4$ possible real roots, then which of the following is true?

• A. None of these
• B. $a<0$
• C. $a=0$
• D. $a>0$

Answer 1

The correct option is B $a<0$

Given that $f\left(x\right)=4x^6+a{x}^3+5x-7=0$

According to Descartes’ Rule of Signs the number of positive real zeros is either equal to the number of sign changes of f(x) or is less than the number of sign changes by an even integer and the number of negative real zeros is either equal to the number of sign changes of $f(-x)$ or is less than the number of sign changes by an even integer.
Let $a>0$,
then in the given $f\left(x\right)$ we have one sign change, so there can be one positive real root.
$f(-x)=4x^6-a{x}^3-5-7$
In the given $f(-x)$ we have one sign change, so there can be one negative real root.
Therefore the total real roots can be $1+1=2$
But it is given that there are $4$ possible real roots.

Let $a<0$,
then in the given $f\left(x\right)$ we have three sign changes, so there can be three positive real root.
$f(-x)=4x^6-a{x}^3-5-7$
Since $a<0$, the given $f(-x)$ will have one sign change, so there can be one negative real root.
Therefore the total real roots can be $3+1=4$
So $a<0$ is the answer.