Solving $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? (1983 AIME problem 3)

Question

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

I know it is a messy/bad idea, but I first started off by squaring both sides and moving everything to one side to get $$x^4 + 36x^3 + 384x^2 + 1080x + 900 - 4x^2 - 72x - 180 = x^4 + 36x^3 + 380x^2 + 1008x + 720 .$$ And by (generalisation) of Vieta's formula, the product of the real roots should be $\frac{720}{1} = 720$, but that is wrong, and I don't understand why.

Answer 1

Use the following substitution. $$x^2+18x+45=t$$ and we obtain $$t-2\sqrt{t}-15=0$$ ot $$(\sqrt{t}-1)^2=16$$ and since $\sqrt{t}\geq0$, we obtain $$\sqrt{t}=5,$$ $$x^2+18x+20=0$$ or $$(x+9)^2=61$$ and we get the answer: $$\{-9\pm\sqrt{61}\}$$

Answer 2

Write $t=x^2+18x +30$, then we get $$t^2=4t+60$$ so $(t-10)(t+6)=0$ and thus

1. case $x^2+18x+36=0$ so $x_1=x_2=-6$ which is impossible (left side is negative)

2. case $x^2+18x+20=0$ so the product of real roots is $20.$

Answer 3

$$x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$$ let $x^2+18x+30=u$ so $$u= 2 \sqrt{u+15}$$ $$u^2=4(u+15)$$ $$u^2-4u-60=0$$ $$(u+6)(u-10)=0$$ so $$({\color{Red} {x^2+18x+36}})({\color{Blue} {x^2+18x+20}})=0$$ the roots of $({\color{Red} {x^2+18x+36}})$ do not satisfy the original equation especially in the term $\sqrt{x^2 + 18x + 45}$ so the product of real roots will be $20$