Difference between revisions of "1983 AIME Problems/Problem 3"

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== See also ==
 
== See also ==
{{AIME box|year=1983|num-b=1|num-a=3}}
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* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]

Revision as of 17:36, 21 March 2007

Problem

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

Solution

If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.

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

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution gives us non-real roots, so we'll will go with the first. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0$. The product of our roots is therefore 20.


See also

1983 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions