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"

(→‎Solution 2)
m (→‎Solution: reworded)
Line 7: Line 7:
 
Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our equation becomes <math>y=2\sqrt{y+15}</math>.
 
Instead, we substitute <math>y</math> for <math>x^2+18x+30</math> and our 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>. The second solution gives us non-real roots (think about it carefully if you don't see why), so we will go with the first. Substituting <math>x^2+18x+30</math> back in for <math>y</math>,
+
Now we can square; solving for <math>y</math>, we get <math>y=10</math> or <math>y=-6</math>. The second solution is extraneous since <math>2\sqrt{y=15}</math> is positive. So, 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> By [[Vieta's formulas]], the product of our roots is therefore <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 our roots is therefore <math>\boxed{020}</math>.
  

Revision as of 21:51, 21 May 2009

Problem

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

Solution

If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.

Instead, we substitute $y$ for $x^2+18x+30$ and our equation becomes $y=2\sqrt{y+15}$.

Now we can square; solving for $y$, we get $y=10$ or $y=-6$. The second solution is extraneous since $2\sqrt{y=15}$ is positive. So, we have $y=10$ as the only solution for $y$. Substituting $x^2+18x+30$ back in for $y$,

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

By Vieta's formulas, the product of our roots is therefore $\boxed{020}$.

Solution 2

There is an error with solution one since the second solution for $y$ does NOT give us non-real roots, since the discriminant is positive. Another way to do this problem would be without substitution, and by just squaring both sides and making the quartic equation equal to 0. We have:

$x^4+36x^3+380x^2+1080x+900 = 0$

By Vieta's formula we get that the product is $\boxed{900}$.

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=31799"