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> | + | 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 (plugging in <math>6</math> as <math>y</math>, we get <math>-</math><math>6</math> <math>=</math> <math>6</math>, which obviously not true). 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 the roots is <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 the roots is <math>\boxed{020}</math>. | ||
Revision as of 18:31, 5 August 2016
Problem
What is the product of the real roots of the equation 
?
Solution
If we expand by squaring, we get a quartic polynomial, which isn't always the easiest thing to deal with.
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 (plugging in 
as , we get 
, which obviously not true). So, we have as the only solution for . Substituting back in for ,
By Vieta's formulas, the product of the roots 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 | ||
