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 is extraneous since <math>2\sqrt{y+15}</math> is positive. So, we have <math>y=10</math> as the only solution for <math>y</math>. 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 is extraneous since <math>2\sqrt{y+15}</math> is positive. So, we have <math>y=10</math> as the only solution for <math>y</math>. Substituting <math>x^2+18x+30</math> back in for <math>y</math>, | ||
<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>. | <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>. | ||
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== See also == | == See also == |
Revision as of 20:14, 22 May 2009
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 is extraneous since is positive. So, we have as the only solution for . Substituting back in for ,
By Vieta's formulas, the product of our roots is therefore .
See also
1983 AIME (Problems • Answer Key • Resources) | ||
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 |