# Multiplying imagionary roots of a polynomial

I am trying to answer the following question:

The roots of the quadratic equation $ax^2-16x+25$ are $2+mi$ and $2-mi$, where $m>0$. Compute the sum of $a+m$.

Should the zeros of the equation be $x-(2+mi)$ and $x-(2-mi)$ or $x-2+mi$ and $x-2-mi$, or maybe something else? I really have no idea.

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It looks like something went wrong... By factoring you get that $ax^2 - 16x = x(ax-16) = 0$, so we have $x = 0$ of $x = \frac{16}{a}$ are the roots. – andybenji Dec 2 '12 at 22:37
@andybenji it looks like someone who edited the problem made a mistake. I fixed it. Can you take another look? – PygameNerd Dec 2 '12 at 22:45
Yeah definitely. – andybenji Dec 2 '12 at 22:46
@andybenji ...and perhaps answer the question as well??? – PygameNerd Dec 2 '12 at 22:49

The roots are given by the quadratic formula: $$x = \frac{16\pm \sqrt{16^2 - 4*25*a}}{2a}$$
We know that the real part of the roots will be $2$, so $\frac{16}{2a} = 2$, and $a = 4$. So we can plug that back into the quadratic formula. $$x = \frac{16\pm \sqrt{16^2 - 4*25*4}}{2*4}$$ and solve for your roots.