Difference between revisions of "1983 AIME Problems/Problem 3"
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<cmath>(x^2+18x+30)^2=4(x^2+18x+30)+60</cmath> | <cmath>(x^2+18x+30)^2=4(x^2+18x+30)+60</cmath> | ||
<cmath>(x^2+18x+30)^2-4(x^2+18x+30)-60=0</cmath> | <cmath>(x^2+18x+30)^2-4(x^2+18x+30)-60=0</cmath> | ||
+ | Substituting <math>y=x^2+18x+30</math> yields | ||
+ | <cmath>y^2-4x-60=0</cmath> | ||
+ | <cmath>(y+6)(y-10)=0</cmath> | ||
+ | Thus <math>y=x^2+18x+30=-6,10</math> | ||
== See Also == | == See Also == |
Revision as of 13:33, 13 August 2019
Contents
[hide]Problem
What is the product of the real roots of the equation ?
Solution
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 .
Solution 3
Begin by completing the square on both sides of the equation, which gives Now by substituting , we get , or The solutions in are then Turns out, are a pair of extraneous solutions. Thus, our answer is then By difference of squares.
Solution 4
We are given the equation Squaring both sides yields Substituting yields Thus
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 |