Difference between revisions of "1983 AIME Problems/Problem 3"
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What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>? | What is the product of the [[real]] [[root]]s of the [[equation]] <math>x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}</math>? | ||
− | == Solution == | + | == Solutions == |
+ | === Solution 1 === | ||
If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with. | If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with. | ||
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<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>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | We begin by noticing that the polynomial on the left is <math>15</math> less than the polynomial under the radical sign. Thus: <cmath>(x^2+ 18 + 45) - 2\sqrt{x^2+18+45} - 15 = 0.</cmath> Letting <math>n = \sqrt{x^2+18+45}</math>, we have <math>n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0</math>. Because the square root of a real number can't be negative, the only possible <math>n</math> is <math>5</math>. | ||
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+ | Substituting that in, we have <cmath>\sqrt{x^2+18+45} = 5 \Longrightarrow x^2 + 18 + 45 = 25 \Longrightarrow x^2+18+20=0.</cmath> | ||
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+ | And by [[Vieta's formulas]], the product of the roots is <math>\boxed{020}</math>. | ||
== See Also == | == See Also == |
Revision as of 10:25, 15 October 2016
Contents
[hide]Problem
What is the product of the real roots of the equation ?
Solutions
Solution 1
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 is 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 .
Solution 2
We begin by noticing that the polynomial on the left is less than the polynomial under the radical sign. Thus: Letting , we have . Because the square root of a real number can't be negative, the only possible is .
Substituting that in, we have
And 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 |