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. $a>0$
• B. None of the above
• C. $a<0$
• D. $a=0$

Answer 1

The correct option is B None of the above

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 $f\left(x\right)$ 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.
$f(-x)=4x^6-|a|x^3-5x-7$
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.

So, now let's take $a<0$,
then in the given $f\left(x\right)$ we have three sign changes, so there can be three positive real roots or one positive real root.
Also, $f(-x)=4x^6-(-a)x^3-5x-7$
$\Rightarrow f(-x)=4x^6+a{x}^3-5x-7$
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 $4$ or $2$
Thus, no real values of $a$ is possible.