Difference between revisions of "1983 AIME Problems/Problem 3"
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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 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>. | ||
Revision as of 18:50, 5 March 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 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 
.
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 | ||
