Difference between revisions of "1983 AIME Problems/Problem 3"
m |
(→Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | If we expand by squaring, we get a quartic [[polynomial]], which | + | If we expand by squaring, we get a quartic [[polynomial]], which isn't very helpful. |
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 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 | + | |
| + | <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>. | ||
== See Also == | == See Also == | ||
Revision as of 14:08, 2 June 2013
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 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 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 | ||
