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 (think about it carefully if you don't see why), so we 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 (think about it carefully if you don't see why), so we will go with the first. 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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| + | == Solution 2 == | ||
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| + | There is an error with solution one since the second solution for <math>y</math> does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have: | ||
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| + | <math>x^4+36x^3+380x^2+1080x+900 = 0</math> | ||
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| + | By Vieta's formula we get that the product is <math>boxed{900}</math>. | ||
== See also == | == See also == | ||
Revision as of 18:56, 20 May 2009
Contents
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 (think about it carefully if you don't see why), so we will go with the first. Substituting back in for ,
By Vieta's formulas, the product of our roots is therefore 
.
Solution 2
There is an error with solution one since the second solution for does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have:
By Vieta's formula we get that the product is 
.
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 | ||
