Common root questions

Question

The value of $a$ for which the equations $x^3+ax+1=0$ and $x^4+ax^2+1=0$ have a common root has to be found.

I tried by subtracting the two equations then I converted them in factors which come out to be $x(x^2+a)(x-1)=0$.

Now what to do?

Answer 1

Compute the gcd with the Euclidean algorithm: \begin{align} x^4+ax^2+1&=x(x^3+ax+1)+(-x+1)\\[4px] x^3+ax+1&=(-x^2-x-a-1)(-x+1)+a+2 \end{align} If $a+2\ne0$, the polynomials are coprime.

Can you finish?

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Alternative method: the resultant is $$ \det\begin{bmatrix} 1 & 0 & a & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & a & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & a & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & a & 1 \\ 1 & 0 & a & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & a & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & a & 0 & 1 \\ \end{bmatrix} =a+2 $$

Answer 2

We have that $$x^4+ax^2+1-x(x^3+ax+1)=1-x.$$ Hence a common root $x$ of the equations $x^3+ax+1=0$ and $x^4+ax^2+1=0$ should solve also the equation $1-x=0$. It follows that the only possible common root is $x=1$. Can you take it form here?

Answer 3

A common root will be a root of the difference - so what are the roots of the difference (they are rather visible now you have factorised it)?

But it will also be a root of the original equations. So which of those roots can solve the original pair of equations (substitute back in)? And what value(s) of $a$ does that imply?

Polynomial division as per egreg's (and partly Robert Z's) answer - the Euclidean Algorithm for polynomials - will find you the highest common factor of the two - it subtracts the right multiple, and isn't just opportunistic. Using that eliminates some of the "false" roots. But your method works fine with care.

Answer 4

Well since you did subtract

If $x^3 + ax + 1= 0 = x^4 + ax^2 + 1 = 0$ then $x^4 - x^3 +a^2x + ax = x(x^2 + a)(1-x)= 0$ so either $x = 0; x=1;$ or $x = \sqrt{-a}$.

If $x = 0$ then $x^3 + ax + 1 = x^4 + ax^2 + 1 = 1 \ne 0$ so that was an extraneous root added.

If $x = 1$ then $x^3 + ax + 1 = a+ 2 = 0$ and $x^4 + ax + 1 = a+2 = 0$. So $a$ must be $a = -2$.

If $x= \sqrt{-a}$ then $x^2 + ax + 1 = -a\sqrt{-a} + a\sqrt{-a} + 1 = 1 \ne 0$ so that was an extraneous root added.

So the only possible common root is $x =1$ and for that to be a common root we must have $a = -2$.

Subtracting added extraneous roots so perhaps there is a more efficient way. No matter. It did work.