Difference between revisions of "1991 AIME Problems/Problem 8"
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
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Revision as of 18:22, 4 July 2013
Problem
For how many real numbers 
does the quadratic equation 
have only integer roots for 
?
Solution
Solution 1
By Vieta's formulas, 
where 
are the roots of the quadratic, and since 
are integers, 
must be an integer. Applying the quadratic formula,
![\[x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}\]](http://latex.artofproblemsolving.com/5/c/9/5c9418734aec5a2653f174c7def6c5e4f5be6df3.png)
Since 
is an integer, we need 
to be an integer (let this be 
): 
. Completing the square, we get
![\[(a - 12)^2 = b^2 + 144\]](http://latex.artofproblemsolving.com/b/c/6/bc6750555824e49ee620d85e9d8f542b7b1a2aa3.png)
Which implies that 
is a perfect square also (let this be 
). Then
![\[c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144\]](http://latex.artofproblemsolving.com/f/e/f/fef068ebf7b69ec2435c17b7aaaf38f76b1e3317.png)
The pairs of factors of 
are 
; since 
is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get 
pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work.
Solution 2
Let 
. Vieta's yields 
.
\begin{eqnarray*}sr + 6s + 6r &=& 0\\
sr + 6s + 6r + 36 &=& 36\\
(s + 6)(r + 6) &=& 36 (Error compiling LaTeX. Unknown error_msg)
Without loss of generality let 
.
The possible values of 
are:
.
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
