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Difference between revisions of "2015 AMC 12A Problems/Problem 18"

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By the quadratic formula, the roots <math>r</math> can be represented by
 
By the quadratic formula, the roots <math>r</math> can be represented by
 
<cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath>
 
<cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath>
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For <math>r\in\mathbb{Z}</math>, <math>a\in\mathbb{Z}</math> (or <math>\frac{\sqrt{a^2-8a}}{2}</math> and \frac{a}{2} will have different mantissae). So we observe the discriminant <math>\sqrt{a^2-8a}=\sqrt{a(a-8)}</math>. We then have two cases.
+
For <math>r\in\mathbb{Z}</math>, <math>a\in\mathbb{Z}</math> (since <math>\frac{\sqrt{a^2-8a}}{2}</math> and <math>\frac{a}{2}</math> will have different mantissae). So we observe the discriminant <math>\sqrt{a^2-8a}=\sqrt{a(a-8)}</math>. We then have two cases.
  
  

Revision as of 21:27, 6 September 2019

Problem

The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$?

$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18$

Solution 1

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.

The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$

As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$, for some nonnegative integer $k$.

$a^2-8a=k^2$

$a(a-8)=k^2$

$((a-4)+4)((a-4)-4)=k^2$

$(a-4)^2-4^2=k^2$

$(a-4)^2=k^2+4^2$

From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$.

In the case $k=0$, the equation simplifies to $|a-4|=4$. From this equation, we have $a=0,8$. For both $a=0$ and $a=8$, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")

If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$. Hence, $a=-1,9$. Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$, so these two values also satisfy the original constraints.

There are a total of four possible values for $a$: $-1,0,8,$ and $9$. Hence, the sum of all of the possible values of $a$ is $\boxed{\textbf{(C) }16}$.


Solution 2 (Quick and Dirty)

By the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$, $a\in\mathbb{Z}$ (since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissae). So we observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$. We then have two cases.


Positive $a$

$a\geq8$ and $a\leq0$, since any $1\geq a \geq7$ yields imaginary roots. Testing positive $a$ values, it quickly becomes clear that $a\leq9$. After $36$ and $49$, the difference between the closest nonzero factor pairs of perfect squares exceeds $8$. In the interval $8\geq a \geq9$, $a=8,9$. Checking $a=9$ yields an integer.

Negative $a$

We can instead test with $\sqrt{b(b+8)}$, where $b=-a$. If $c=b+8$, we have our original discriminant. For the same reasons, $z=8,9\implies 8,9=8-a$. $a=-1$ (0 also works but does not affect the answer).


$-1+8+9=16\implies\boxed{\textbf{(C) }16}$

Solution 3

Let $m$ and $n$ be the roots of $x^2-ax+2a$

By Vieta's Formulas, $n+m=a$ and $mn=2a$

Substituting gets us $n+m=\frac{mn}{2}$

$2n-mn+2m=0$

Using Simon's Favorite Factoring Trick:

$n(2-m)+2m=0$

$-n(2-m)-2m=0$

$-n(2-m)-2m+4=4$

$(2-n)(2-m)=4$

This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$. Adding these up gets $\boxed{\textbf{(C) }16}$.

Solution 4

The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\]. For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]

We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$, we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$. These are the only ones; if we want to make sure then we must hand check up to $b=8$. Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find 16 (C) as the answer.

Additional note: You can use the quadratic and plug in squares for a (since for b^2 to be an integer a would have to be some square), and eventually you can notice a limit to get the answer~

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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