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

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== Problem ==
 
== Problem ==
What is the [[product]] of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
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What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>?
  
 
== Solution ==
 
== Solution ==
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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>.
 
Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.
  
Now we can square; 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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Now we can square; 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 will go with the first. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,
 
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<center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> By [[Vieta's formulas]], the product of our roots is therefore <math>\boxed{020}</math>.
<math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0</math>. The product of our roots is therefore 20.
 
 
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1983|num-b=2|num-a=4}}
 
{{AIME box|year=1983|num-b=2|num-a=4}}
 
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 10:35, 25 April 2008

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 will go with the first. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$

By Vieta's formulas, the product of our roots is therefore $\boxed{020}$.

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