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 $\boxed{20}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1983 AIME Problems/Problem 3

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