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I got an question to find the product of the roots of $x^2 + 18x + 30 = 2\sqrt{x^2 + 18x + 45}$. This is what I did:
$y = x^2 + 18x$. So $y + 30 = 2\sqrt{y + 45}$.

Squaring it on both sides:

$(y + 30)^2 = (2\sqrt{y + 45})^2\Rightarrow y^2 + 30^2 + 60y = 4(y + 45)\Rightarrow y^2 + 900 + 60y - 4y - 180 = 0\Rightarrow y^2 + 720 + 56y = 0$.

Then using quadratic formula I got the values of $y$ as $-20$ and $-36$. Now plugging In the values as $x^2 + 18x + 36/+20 = 0$, I don't get a whole number as the value for $x$. This is the place I'm stuck.

Any help is appreciated

infixint943
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  • Stuck? Why? Your equation has no integer solutions. – José Carlos Santos Nov 07 '17 at 16:42
  • I mean I don't know how to find the product of the roots after that step – infixint943 Nov 07 '17 at 16:47
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    If $f(x)$ is any polynomial at all, what is the product of its roots? If you aren't familiar with Vieta's formulas, just think about a factorization of $f(x) = \prod (x-a_i)$ where the $a_i$ are the roots. What is $\prod a_i$ in terms of the coefficients of $f$? – rogerl Nov 07 '17 at 16:49
  • Product of solutions is $(-9+\sqrt{61})(-9-\sqrt{61})=81-61=20$ The other equation gives solutions which are not actual solutions of the given equation. They come from the squaring and are extraneous (how do you call them ?) – Raffaele Nov 07 '17 at 17:39

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The product of the (real or complex) roots of $x^2+18x+36$ is $36$, the product of those of $x^2+18x+20$ is $20$, hence the product of the roots of $(x^2+18x+36)(x^2+18x+20)$ is $720$.

There would remain to eliminate the roots which do not belong to the domain of validity of the initial equation, the roots of which must satisfy: $$x^2+18x+30\ge0.$$

Bernard
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