Difference between revisions of "1983 AIME Problems/Problem 3"
Sevenoptimus (talk | contribs) m (Changed "Solutions" to "Solution" (to match other AIME pages)) |
(Solution 3) |
||
Line 18: | Line 18: | ||
Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>. | Reasoning as in Solution 1, the product of the roots is <math>\boxed{020}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Begin by completing the square on both sides of the equation, which gives <cmath>(x+9)^2-51=2\sqrt{(x+3)(x+15)}</cmath> | ||
+ | Now by substituting <math>y=x+9</math>, we get <math>y^2-51=2\sqrt{(y-6)(y+6)}</math>, or <cmath>y^4-106y^2+2745=0</cmath> | ||
+ | The solutions in <math>y</math> are then <cmath>y=x+9=\pm3\sqrt{5},\pm\sqrt{61}</cmath> | ||
+ | Turns out, <math>\pm3\sqrt{5}</math> are a pair of extraneous solutions. Thus, our answer is then <cmath>\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}</cmath> | ||
+ | By difference of squares. | ||
== See Also == | == See Also == |
Revision as of 00:14, 3 May 2019
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.
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 |