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

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We want <math>16 + b^2</math> to be a perfect square. From smartly trying small values of <math>b</math>, we find <math>b = 0, b = 3</math> as solutions, which correspond to <math>a = -1, 0, 8, 9</math>. These are the only ones; if we want to make sure then we must hand check up to <math>b=8</math>. Indeed, for <math>b \geq 9</math> we have that the differences between consecutive squares are greater than <math>16</math> so we can't have <math>b^2 + 16</math> be a perfect square. So summing our values for <math>a</math> we find <math>\boxed{16 \textbf{ (C)}}</math>. as the answer.
 
We want <math>16 + b^2</math> to be a perfect square. From smartly trying small values of <math>b</math>, we find <math>b = 0, b = 3</math> as solutions, which correspond to <math>a = -1, 0, 8, 9</math>. These are the only ones; if we want to make sure then we must hand check up to <math>b=8</math>. Indeed, for <math>b \geq 9</math> we have that the differences between consecutive squares are greater than <math>16</math> so we can't have <math>b^2 + 16</math> be a perfect square. So summing our values for <math>a</math> we find <math>\boxed{16 \textbf{ (C)}}</math>. 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~
+
Additional note: You can use the quadratic and plug in squares for a (since for <math>b^2</math> to be an integer <math>a</math> would have to be some square), and eventually you can notice a limit to get the answer~
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2015|ab=A|num-b=17|num-a=19}}

Revision as of 12:40, 1 August 2021

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

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 mantissas (mantissae?).

Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.


Positive $a$

$a\geq8$ and $a\leq0$, since $1\geq a \geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\leq9$. After $16$ and $36$, the difference between the closest nonzero factor pairs of perfect squares exceeds $8$. For $8\geq a \geq9$, $a=8,9$. Checking both yields an integer.

Negative $a$

We can instead test with $\sqrt{-a(8-a)}$. If $b=8-a$, we have our original expression. Thus, for the same reasons, $b=8,9\implies 8,9=8-a$. $a=-1$ (0 does not affect the answer).


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


(Solution by BJHHar)

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 $\boxed{16 \textbf{ (C)}}$. 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 (ProblemsAnswer KeyResources)
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