Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1983 AIME Problems/Problem 3"

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 01:14, 3 May 2019

Problem

What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?

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 $y$ for $x^2+18x+30$, so that the equation becomes $y=2\sqrt{y+15}$.

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$, we get $-6=6$, which is obviously false). Hence we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$

Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$, which is positive. Thus by Vieta's formulas, the product of the real roots is simply $\boxed{020}$.

Solution 2

We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18x+45}$, we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$. Because the square root of a real number can't be negative, the only possible $n$ is $5$.

Substituting that in, we have \[\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.\]

Reasoning as in Solution 1, the product of the roots is $\boxed{020}$.

Solution 3

Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$, we get $y^2-51=2\sqrt{(y-6)(y+6)}$, or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \[\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}\] 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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_3&oldid=105597"