If one real root of 81x^2+kx+256=0 is the cube of the other root, a value of k is:

Question

If one real root of $81x^2+kx+256=0$ is the cube of the other root, a value of $k$ is:

Accepted Answer

Let roots be $\beta,\beta^3$. Product $=\beta^4=\dfrac{256}{81} \implies \beta=\dfrac43$, so the other root is $\dfrac{64}{27}$.

Sum $=\dfrac43+\dfrac{64}{27}=-\dfrac k{81} \implies k=-300.$

Solutions · 1