1983 AIME Problems/Problem 3

Problem

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

Solutions

Solution 1

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 is 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 $\boxed{020}$ .

Solution 2

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: $(x^2+ 18 + 45) - 2\sqrt{x^2+18+45} - 15 = 0.$ Letting $n = \sqrt{x^2+18+45}$ , we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$ .

Substituting that in, we have $\sqrt{x^2+18+45} = 5 \Longrightarrow x^2 + 18 + 45 = 25 \Longrightarrow x^2+18+20=0.$

And by Vieta's formulas, the product of the roots is $\boxed{020}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1983 AIME Problems/Problem 3

Contents

Problem

Solutions

Solution 1

Solution 2

See Also