1983 AIME Problems/Problem 3

Revision as of 19:31, 5 August 2016 by M1234567 (talk | contribs)

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 isn't always the easiest thing to deal with.

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 (plugging in $6$ as $y$, we get $-$$6$ $=$ $6$, which obviously not true). 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 the roots is $\boxed{020}$.

See Also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions