Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 4)
(→‎Solution 6 (big fast))
 
(3 intermediate revisions by 2 users not shown)
Line 114: Line 114:
  
  
−
We want the possible values of <math>m</math>, which are <math>-5,-4,4,</math> and <math>5</math>. As <math>m+4=a</math>, <math>a</math> can equal <math>-1,0,8,</math> or <math>9.</math> Adding all of that up gets us our answer, <math>\boxed{16 \textbf{ (C)}}</math>.
+
We want the possible values of <math>m</math>, which are <math>-5,-4,4,</math> and <math>5</math>. As <math>m+4=a</math>, <math>a</math> can equal <math>-1,0,8,</math> or <math>9.</math> Adding all of that up gets us our answer, <math>\boxed{\textbf{(C) }16}</math>.
 +
 
 +
 
 +
(Solution by Curious_crow)
 +
 
 +
== Solution 6 (big fast) ==
 +
Let our roots be <math>r</math> and <math>s</math>. By vieta's formulas, we have that <math>r+s=a</math> and <math>rs = 2a</math>. Thus, <math>rs = 2r+2s</math>, or <math>(r-2)(s-2) = 4</math>.
 +
 
 +
4 can only be obtained from (2,2), (1,4), (-2,-2), or (-1,-4), giving us <math>-1,0,8</math> and <math>9</math>, which add up to <math>\boxed{\textbf{(C)}16}</math>
 +
 
 +
-skibbysiggy
  
 
== 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}}

Latest revision as of 23:19, 16 September 2024

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~


Solution 5

First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.

We can complete the square and rearrange to get $(a-4)^2-b^2=16$. Let's define $m=a-4$, just to make things a little easier to write, so now we have $(m+b)(m-b)=16$. We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$. (Note that $m$ and $b$ must both be integers)


$(m+b)(m-b)=16$

$16*1=16$ $\Rightarrow$ Doesn't work

$8*2=16$ $\Rightarrow$ $m=5, b=3$

$4*4=16$ $\Rightarrow$ $m=4, b=0$

$2*8=16$ $\Rightarrow$ $m=5, b=-3$

$1*16=16$ $\Rightarrow$ Doesn't work

$-16*-1=16$ $\Rightarrow$ Doesn't work

$-8*-2=16$ $\Rightarrow$ $m=-5, b=-3$

$-4*-4=16$ $\Rightarrow$ $m=-4, b=0$

$-2*-8=16$ $\Rightarrow$ $m=-5, b=-3$

$-1*-16=16$ $\Rightarrow$ Doesn't work


We want the possible values of $m$, which are $-5,-4,4,$ and $5$. As $m+4=a$, $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\boxed{\textbf{(C) }16}$.


(Solution by Curious_crow)

Solution 6 (big fast)

Let our roots be $r$ and $s$. By vieta's formulas, we have that $r+s=a$ and $rs = 2a$. Thus, $rs = 2r+2s$, or $(r-2)(s-2) = 4$.

4 can only be obtained from (2,2), (1,4), (-2,-2), or (-1,-4), giving us $-1,0,8$ and $9$, which add up to $\boxed{\textbf{(C)}16}$

-skibbysiggy

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_18&oldid=227376"