# Two roots of $4x^3+8x^2+Kx-18=0$ are equal numerically but opposite in value. Find the value of K.

Two roots of $4x^3+8x^2+Kx-18=0$ are equal numerically but opposite in value. Find the value of K.

I've tried plugging in two roots $x_0$ and $-x_0$, to get the relationship $k=-4x^2$. Plugging it back in doesn't achieve anything, I'm not sure how to proceed, I have seen people do this with the quadratic, but similarly trying to force this answer is too complicated, this is an SAT II Level 2 question and should be solvable in under a minute!

If $a$ is one of the two opposite roots, then \begin{align} (x^2-a^2)(4x-r)&=4x^3+8x^2+Kx-18\\ 4x^3-rx^2-4a^2x+a^2r&=4x^3+8x^2+Kx-18 \end{align} The coefficient of $x^2$ reveals the value of $r$. The constant term then reveals the value of $a^2$. And then the linear coefficient reveals the value of $K$.
• So the answer is $-12$. Nice! – MaximusFastidiousIrreverence Apr 22 '18 at 4:35
Hint: by Vieta's relations the sum of all three roots is $-8/4=-2\,$. Now, if two of them sum to $0$ $\ldots$
Let the roots be $a, - a, b$. Then
$$4x^3+8x^2+Kx-18=4(x^2-a^2)(x-b)$$
So, $-4b=8$, $-4a^2=K$ and $4a^2b=-18$.