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

 
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== Problem ==
 
== Problem ==
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What is the product of the real roots of the equation <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
  
 
== Solution ==
 
== Solution ==
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If we expand this, we get a quartic [[polynomial]], which obviously isn't very helpful.
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Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.
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Solving for <math>y</math>, we get <math>y=10</math> or <math>y=-6</math>. The second solution gives us non-real roots, so we'll will go with the first. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,
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<math>x^2+18x+30=10 \Rightarrow x^2+18x+20=0</math>. The product of our roots is therefore 20.
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----
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* [[1983 AIME Problems/Problem 2|Previous Problem]]
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* [[1983 AIME Problems/Problem 4|Next Problem]]
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* [[1983 AIME Problems|Back to Exam]]
  
 
== See also ==
 
== See also ==
* [[1983 AIME Problems]]
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* [[AIME Problems and Solutions]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 22:53, 23 July 2006

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 this, we get a quartic polynomial, which obviously isn't very helpful.

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

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 $y$,

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


See also