# 1983 AIME Problems/Problem 3

## Problem

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?

## Solution

If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.

Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\sqrt{y+15}$.

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution is extraneous since $2\sqrt{y+15}$ is positive. So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$

By Vieta's formulas, the product of our roots is therefore $\boxed{020}$.

## Solution 2

There is an error with solution one since the second solution for $y$ does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have:

$x^4+36x^3+380x^2+1080x+900 = 0$

By Vieta's formula we get that the product is $\boxed{900}$.