Number of values of a for common roots

Question

If $x^3+ax+1=0$ and $x^4+ax^2+1=0$ have a common root, then the number of values of '$a$' are.......

I tried using $x=p$, and then trying to do some manupilation but no result achieved.

Answer 1

Note that$$x^4+ax^2+1=x(x^3+ax+1)-x+1.$$So, if your polynomials have a common root $r$, then $r$ will also be a root of $-x+1$. Can you do the rest?

Answer 2

$0$ is not a root and $$x^4+ax^2+x=0,$$ which gives that $1$ is an unique common root and from here $a=-2.$

Answer 3

Equating both equation $$x^4+ax^2+1=x^3+ax+1\\ x(x-1)(x^2+a)=0\\ x=0,1,\sqrt{-a}$$ One of these $ 0,1,\sqrt{-a}\;$ must be the common root of both equation. $x=0,\sqrt{-a}$ are not the roots of equation. So $x=1\implies a=-2$