One of the roots of $x^2-m^4x+m^2-m^4=0$ is $\sqrt[3]2-\sqrt[3]4$. What is the value of $m$ and other root?

Question

If one of the solutions of $x^2-m^4x+m^2-m^4=0$ be equal to $\sqrt[3]2-\sqrt[3]4$, find $m$ and other root of the equation.

To solve this problem I tried different approaches (like plugging $x=\sqrt[3]2-\sqrt[3]4$ in the equation or using quadratic formula $\frac{-b\pm\sqrt{\Delta}}{2a}$) but neither of them worked. Finally I realized that if we assume the other root is equal to conjugate of the root ($x_2=\sqrt[3]2+\sqrt[3]4$) we have $m^4=2\sqrt[3]2\;\Rightarrow m=\pm\sqrt[3]2$. And we can check product of roots is equal to $m^2-m^4$ (i.e $\sqrt[3]4-\sqrt[3]16)$.

My question is, in general if $\sqrt[n]p -\sqrt[n]q$ (for $n\in\mathbb{N}$) is one of a root of a quadratic equation, can we conclude the other root should be $\sqrt[n]p+\sqrt[n]q$ or it happened here by chance?

Furthermore, I'll appreciate other approaches for solving the problem.

Answer 1

Another way to look at it... the constant term $m^2 - m^4$ factors into $(m - m^2)(m + m^2)$. Since it's the product of the two roots and $\sqrt[3]2 - \sqrt[3]4$ is one of the roots, it would make sense that $m =\sqrt[3]2$ here and $\sqrt[3]2 + \sqrt[3]4$ is the other root.

To verify it, the $x$ coefficient would have to be $-1$ times the sum of the two roots, or $-2\sqrt[3]2$. Hence you'd have to have $-m^4 = -2\sqrt[3]2$, which does hold for $m = \sqrt[3]2$. So you have the correct two roots.

Answer 2

The following answers this part of the question.

My question is, in general if $\sqrt[n]p -\sqrt[n]q$ (for $n\in\mathbb{N}$) is one of a root of a quadratic equation, can we conclude the other root should be $\sqrt[n]p+\sqrt[n]q$or it happened here by chance?

No, a general quadratic $\,\big(x-r\big)\big(x - \left(\sqrt[n]p -\sqrt[n]q\right)\big)\,$ has roots $\,\sqrt[n]p -\sqrt[n]q\,$ and $\,r\,$ for arbitrary $\,r\,$.

This leaves the question open for the particular quadratic considered here, with coefficients of the given form as functions of $\,m\,$. The answer is still negative, and it turns out that $\,\sqrt[3]{2}-\sqrt[3]{4}\,$ is a special case that was specifically chosen for $\,\sqrt[3]{2}+\sqrt[3]{4}\,$ to also be a root.

Suppose the quadratic $\,x^2-m^4x+m^2-m^4\,$ has a root $\,x = a - a^2\,$. The conditions for the other root to be $\,x = a + a^2\,$ are:

$$ \begin{align} m^4 &= (a-a^2)+(a+a^2)=2a \tag{1} \\ m^2-m^4 &= (a-a^2)(a+a^2) = a^2 - a^4 \tag{2} \end{align} $$

Writing $\,(2)\,$ as $\,m^2 = a^2-a^4+m^4\,$, squaring, and substituting $\,m^4=2a\,$ from $\,(1)\,$:

$$ 0 = (a^2-a^4+m^4)^2-2a=a (a^3 - 2) (a^4 - 2 a^2 - 2 a + 1) $$

Other than the trivial solution $\,a=0\,$, the real solutions are $\,a=\sqrt[3]{2}\,$ and two more real roots of the quartic factor (not pretty to calculate).

In other words, the only integer $\,n\,$ such that both $\,a\pm a^2=\sqrt[3]{n}\pm\sqrt[3]{n^2}\,$ are roots of the given quadratic is $\,n=2\,$. It so "happens" that $\,n=2\,$ is precisely the value used in OP's problem.