# 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

### Solution 1

If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.

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

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$, we get $-6=6$, which is obviously false). Hence we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

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

Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$, which is positive. Thus by Vieta's formulas, the product of the real roots is simply $\boxed{020}$.

### Solution 2

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: $$(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.$$ Letting $n = \sqrt{x^2+18x+45}$, 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 $$\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.$$

Reasoning as in Solution 1, the product of the roots is $\boxed{020}$.

### Solution 3

Begin by completing the square on both sides of the equation, which gives $$(x+9)^2-51=2\sqrt{(x+3)(x+15)}$$ Now by substituting $y=x+9$, we get $y^2-51=2\sqrt{(y-6)(y+6)}$, or $$y^4-106y^2+2745=0$$ The solutions in $y$ are then $$y=x+9=\pm3\sqrt{5},\pm\sqrt{61}$$ Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then $$\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}$$ By difference of squares.

## Solution 4

We are given the equation $$x^2+18x+30=2\sqrt{x^2+18x+45}$$ Squaring both sides yields $$(x^2+18x+30)^2=4(x^2+18x+45)$$ $$(x^2+18x+30)^2=4(x^2+18x+30+15)$$ $$(x^2+18x+30)^2=4(x^2+18x+30)+60$$ $$(x^2+18x+30)^2-4(x^2+18x+30)-60=0$$ Substituting $y=x^2+18x+30$ yields $$y^2-4y-60=0$$ $$(y+6)(y-10)=0$$ Thus $y=x^2+18x+30=-6,10$. However if $y=-6$, the left side of the equation $$x^2+18x+30=2\sqrt{x^2+18x+45}$$ would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have $$x^2+18x+30=10$$ $$x^2+18x+20=0$$ Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$.

~ Nafer