Find the ordered pair $(a,b)$ of real numbers for which $x^2+ax+b$ has a non-real root whose cube is $343$.

Question

Because $x^2 + ax + b$ has all real coefficients, we know the roots must be complex conjugates. So let $\lambda e^{i \theta}$ be a root, so we know that $(x - \lambda e^{i \theta})(x - \lambda e^{-i \theta}) = x^2 - \lambda x (e^{i \theta} + e^{-i\theta}) + \lambda^2$ is our polynomial.

We also know that $(\lambda e^{i \theta})^3 = 7^3 \implies \lambda e^{i \theta} = 7$, and since $|\lambda e^{i \theta}| = |\lambda| = 7,$ we know that $\lambda = 7$.

This means that $e^{i\theta} = 1$ and so $\theta = 0$. But clearly this doesn't give the right answer, since our polynomial isn't supposed to have real roots. Where is the flaw in my reasoning?

Answer 1

$(\lambda e^{i \theta})^3 = 7^3 \implies e^{3i \theta} = (\frac{7}{\lambda})^3e^{0}$

Using De Moivre's theorem

$\implies e^{i \theta} = (\frac{7}{\lambda}) e^{\frac{2k\pi}{3}i}, \text{k = 0,1,2}$

$\implies \lambda=7$

But as given in the question, roots are not real, so $k$ can not be $0$, therefore,

Roots are $7e^{\frac{2\pi}{3}i}$ and $7e^{\frac{4\pi}{3}i}$

$\implies 7(-\frac{1}{2} + \frac{\sqrt3}{2}i) \text{ and } 7(-\frac{1}{2} - \frac{\sqrt3}{2}i)$

Answer 2

We have $x^3 = 343$ because we need to find an $x$ that when cubed is $343$. Now rearrange to form $x^3 - 343=0$. Use the difference of cubes factorization which is $a^3-b^3=(a-b)(a^2+ab+b^2)$ by letting $a=x$ and $b=7$. So, we get $(x-7)(x^2+7x+49) = 0$. This simplifies down to $(x^2+7x+49) = 0$, which is a quadratic with roots that fit our condition that $x^3$ = $343$, because we derived this equation from that fact. Use the quadratic formula on $x^2+7x+49$ to get $\frac{-7}2 \pm {7i\sqrt{3} \over 2}$ as our nonreal roots that when cubed make $343$.