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Difference between revisions of "1991 AIME Problems/Problem 8"
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== Problem ==
 
== Problem ==
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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>?
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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 ===
+
== 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\\
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(s + 6)(r + 6) &=& 36
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\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]],  
  
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<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.
−
 
 
−
=== Solution 2 ===
 
−
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</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>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 22:55, 25 July 2024

Problem

For how many real numbers $a$ does the quadratic equation $x^2 + ax + 6a=0$ have only integer roots for $x$?

Solution 1

Let $x^2 + ax + 6a = (x - s)(x - r)$. Vieta's yields $s + r = - a, sr = 6a$. \begin{eqnarray*}sr + 6s + 6r &=& 0\\ sr + 6s + 6r + 36 &=& 36\\ (s + 6)(r + 6) &=& 36 \end{eqnarray*}

Without loss of generality let $r \le s$.

The possible values of $(r + 6,s + 6)$ are: $( - 36, - 1),( - 18, - 2),( - 12, - 3),( - 9, - 4),( - 6, - 6),(1,36),(2,18),(3,12),(4,9),(6,6)$ $\Rightarrow \boxed{10}\ \text{values of } a$.

Solution 2

By Vieta's formulas, $x_1 + x_2 = -a$ where $x_1, x_2$ are the roots of the quadratic, and since $x_1,x_2$ are integers, $a$ must be an integer. Applying the quadratic formula,

\[x = \frac{-a \pm \sqrt{a^2 - 24a}}{2}\]

Since $-a$ is an integer, we need $\sqrt{a^2-24a}$ to be an integer (let this be $b$): $b^2 = a^2 - 24a$. Completing the square, we get

\[(a - 12)^2 = b^2 + 144\]

Which implies that $b^2 + 144$ is a perfect square also (let this be $c^2$). Then

\[c^2 - b^2 = 144 \Longrightarrow (c+b)(c-b) = 144\]

The pairs of factors of $144$ are $(\pm1,\pm144),( \pm 2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)$; since $c$ is the average of each respective pair and is also an integer, the pairs that work must have the same parity. Thus we get $\boxed{10}$ 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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