Difference between revisions of "1991 AIME Problems/Problem 8"
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Let <math>x^2 + ax + 6a = (x - s)(x - r)</math>. Vieta's yields <math>s + r = - a, sr = 6a</math>. | Let <math>x^2 + ax + 6a = (x - s)(x - r)</math>. Vieta's yields <math>s + r = - a, sr = 6a</math>. | ||
<cmath>\begin{eqnarray*}sr + 6s + 6r &=& 0\\ | <cmath>\begin{eqnarray*}sr + 6s + 6r &=& 0\\ | ||
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<math>( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)</math> | <math>( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)</math> | ||
<math>\Rightarrow \boxed{10}\ \text{values of } a</math>. | <math>\Rightarrow \boxed{10}\ \text{values of } a</math>. | ||
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+ | == Solution 2 == | ||
+ | By [[Vieta's formulas]], <math>x_1 + x_2 = -a</math> where <math>x_1, x_2</math> are the roots of the quadratic, and since <math>x_1,x_2</math> are integers, <math>a</math> must be an integer. Applying the [[quadratic formula]], | ||
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+ | <cmath>x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}</cmath> | ||
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+ | Since <math>-a</math> is an integer, we need <math>\sqrt{a^2-24a}</math> to be an integer (let this be <math>b</math>): <math>b^2 = a^2 - 24a</math>. [[Completing the square]], we get | ||
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+ | <cmath>(a - 12)^2 = b^2 + 144</cmath> | ||
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+ | Which implies that <math>b^2 + 144</math> is a [[perfect square]] also (let this be <math>c^2</math>). Then | ||
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+ | <cmath>c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144</cmath> | ||
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+ | The pairs of factors of <math>144</math> are <math>(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since <math>c</math> is the [[mean|average]] of each respective pair and is also an integer, the pairs that work must have the same [[parity]]. Thus we get <math>\boxed{10}</math> pairs (counting positive and negative) of factors that work, and substituting them backwards show that they all work. | ||
== See also == | == See also == |
Latest revision as of 23:55, 25 July 2024
Problem
For how many real numbers does the quadratic equation
have only integer roots for
?
Contents
Solution 1
Let . Vieta's yields
.
Without loss of generality let .
The possible values of are:
.
Solution 2
By Vieta's formulas, where
are the roots of the quadratic, and since
are integers,
must be an integer. Applying the quadratic formula,
Since is an integer, we need
to be an integer (let this be
):
. Completing the square, we get
Which implies that is a perfect square also (let this be
). Then
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.
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 |
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