Difference between revisions of "1991 AIME Problems/Problem 8"
(+ solution 2, revised version of 1234567890's) |
m (→Solution) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | For how many real numbers <math>a | + | For how many real numbers <math>a</math> does the [[quadratic equation]] <math>x^2 + ax + 6a=0</math> have only integer roots for <math>x</math>? |
__TOC__ | __TOC__ | ||
− | == Solution == | + | |
− | === | + | == Solution 1== |
+ | |||
+ | 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\\ | ||
+ | sr + 6s + 6r + 36 &=& 36\\ | ||
+ | (s + 6)(r + 6) &=& 36 | ||
+ | \end{eqnarray*} | ||
+ | </cmath> | ||
+ | |||
+ | [[Without loss of generality]] let <math>r \le s</math>. | ||
+ | |||
+ | The possible values of <math>(r + 6,s + 6)</math> are: | ||
+ | <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>. | ||
+ | |||
+ | == 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]], | 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]], | ||
Line 17: | Line 32: | ||
<cmath>c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144</cmath> | <cmath>c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144</cmath> | ||
− | 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 | + | 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 == | ||
Line 35: | Line 38: | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.