Difference between revisions of "1983 AIME Problems/Problem 3"
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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'll will go with the first. Substituting <math>x^2+18x+30</math> back in for <math>y</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>, | ||
| − | <math>x^2+18x+30=10 \ | + | <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=1|num-a=3}} | ||
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* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
Revision as of 16:36, 21 March 2007
Problem
What is the product of the real roots of the equation 
?
Solution
If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.
Instead, we substitute 
for 
and our equation becomes 
.
Now we can square; solving for , we get 
or 
. The second solution gives us non-real roots, so we'll will go with the first. Substituting back in for ,
. The product of our roots is therefore 20.
See also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
