AIME 1983 · 第 3 题

Solution 1

Let $y=x^2+18x+30$. Substituting into the original equation yields $y=2\sqrt{y+15}$.

Squaring both sides yields $y^2 = 4y + 60$, which can be rearranged to form $y^2 - 4y - 60 = (y-10)(y+6) = 0$. Though this quadratic has roots $-6$ and $10$, only $y=10$ satisfies the original equation.

So $y=x^2+18x+30=10$, and therefore $x^2+18x+20=0$. This quadratic has discriminant $18^2 - 4 \cdot 1 \cdot 20 = 244 > 0$, so both its roots are real. By Vieta's Formulas, the product of these 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:

$$(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.$$ 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, $\boxed{020}$.

~ Nafer