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 $y^2-4x-60=0$ $(y+6)(y-10)=0$ 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 $x^2+18x+30=10$ $x^2+18x+20=0$ By

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