Difference between revisions of "1983 AIME Problems/Problem 3"
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== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | If we expand by squaring, we get a quartic [[polynomial]], which isn't always the easiest thing to deal with. | + | If we were to expand by squaring, we would get a quartic [[polynomial]], which isn't always the easiest thing to deal with. |
− | Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> | + | Instead, we substitute <math>y</math> for <math>x^2+18x+30</math>, so that the 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 root is extraneous since <math>2\sqrt{y+15}</math> is always non-negative (and moreover, plugging in <math>y=-6</math>, we get <math>-6=6</math>, which is obviously false). Hence 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> | + | <center><math>x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.</math></center> Both of the roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's formulas]], the product of the real roots is simply <math>\boxed{020}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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Substituting that in, we have <cmath>\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.</cmath> | Substituting that in, we have <cmath>\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.</cmath> | ||
− | + | Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>. | |
== See Also == | == See Also == |
Revision as of 18:08, 15 February 2019
Contents
[hide]Problem
What is the product of the real roots of the equation ?
Solutions
Solution 1
If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.
Instead, we substitute for , so that the equation becomes .
Now we can square; solving for , we get or . The second root is extraneous since is always non-negative (and moreover, plugging in , we get , which is obviously false). Hence we have as the only solution for . Substituting back in for ,
Both of the roots of this equation are real, since its discriminant is , which is positive. Thus by Vieta's formulas, the product of the real roots is simply .
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
Reasoning as in Solution 1, 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 |