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

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

@@ Line 3: / Line 3: @@
 The zeros of the function <math>f(x) = x^2-ax+2a</math> are integers. What is the sum of the possible values of <math>a</math>?
-<math> \textbf{(A) }7\qquad\textbf{(B) }8\qquad\textbf{(C) }16\qquad\textbf{(D) }17\qquad\textbf{(E) }18</math>
+<math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18</math>
 ==Solution 1==
@@ Line 31: / Line 31: @@
-==Solution 2 (Quick and Dirty)==
+==Solution 2 ==
 By the quadratic formula, the roots <math>r</math> can be represented by
 <cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath>
-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 mantissas (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 mantissas (mantissae?).
+Now observe the discriminant <math>\sqrt{a^2-8a}=\sqrt{a(a-8)}</math> and have two cases.
 '''Positive <math>a</math>'''
-<math>a\geq8</math> and <math>a\leq0</math>, since any <math>1\geq a \geq7</math> yields imaginary roots. Testing positive <math>a</math> values, it quickly becomes clear that <math>a\leq9</math>. After <math>36</math> and <math>49</math>, the difference between the closest nonzero factor pairs of perfect squares exceeds <math>8</math>. In the interval <math>8\geq a \geq9</math>, <math>a=8,9</math>. Checking <math>a=9</math> yields an integer.
+<math>a\geq8</math> and <math>a\leq0</math>, since <math>1\geq a \geq7</math> gives imaginary roots. Testing positive <math>a</math> values, quickly see that <math>a\leq9</math>. After <math>16</math> and <math>36</math>, the difference between the closest nonzero factor pairs of perfect squares exceeds <math>8</math>. For <math>8\geq a \geq9</math>, <math>a=8,9</math>. Checking both yields an integer.
 '''Negative <math>a</math>'''
-We can instead test with <math>\sqrt{b(b+8)}</math>, where <math>b=-a</math>. If <math>c=b+8</math>, we have our original discriminant. For the same reasons, <math>z=8,9\implies 8,9=8-a</math>. <math>a=-1</math> (0 also works but does not affect the answer).
+We can instead test with <math>\sqrt{-a(8-a)}</math>. If <math>b=8-a</math>, we have our original expression. Thus, for the same reasons, <math>b=8,9\implies 8,9=8-a</math>. <math>a=-1</math> (0 does not affect the answer).
 <math>-1+8+9=16\implies\boxed{\textbf{(C) }16}</math>
+(Solution by BJHHar)
 ==Solution 3==
@@ Line 74: / Line 79: @@
 The quadratic formula gives <cmath>x = \frac{a \pm \sqrt{a(a-8)}}{2}</cmath>. For <math>x</math> to be an integer, it is necessary (and sufficient!) that <math>a(a-8)</math> to be a perfect square. So we have <math>a(a-8) = b^2</math>; this is a quadratic in itself and the quadratic formula gives <cmath>a = 4 \pm \sqrt{16 + b^2}</cmath>
-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 '''16 (C)''' 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 <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~
+==Solution 5==
+First of all, we know that <math>a</math> is the sum of the quadratic's two roots, by Vieta's formulas. Thus, <math>a</math> must be an integer. Then, we notice that the discriminant <math>a^2-8a</math> must be equal to a perfect square so that the roots are integers. Thus, <math>a(a-8)=b^2</math> where <math>b</math> is an integer.
+We can complete the square and rearrange to get <math>(a-4)^2-b^2=16</math>. Let's define <math>m=a-4</math>, just to make things a little easier to write, so now we have <math>(m+b)(m-b)=16</math>. We can now list out the integer factor pairs of 16 and the resulting values of <math>m</math> and <math>b</math>. (Note that <math>m</math> and <math>b</math> must both be integers)
+<math>(m+b)(m-b)=16</math>
+<math>16*1=16</math> <math>\Rightarrow</math> Doesn't work
+<math>8*2=16</math> <math>\Rightarrow</math> <math>m=5, b=3</math>
+<math>4*4=16</math> <math>\Rightarrow</math> <math>m=4, b=0</math>
+<math>2*8=16</math> <math>\Rightarrow</math> <math>m=5, b=-3</math>
+<math>1*16=16</math> <math>\Rightarrow</math> Doesn't work
+<math>-16*-1=16</math> <math>\Rightarrow</math> Doesn't work
+<math>-8*-2=16</math> <math>\Rightarrow</math> <math>m=-5, b=-3</math>
+<math>-4*-4=16</math> <math>\Rightarrow</math> <math>m=-4, b=0</math>
+<math>-2*-8=16</math> <math>\Rightarrow</math> <math>m=-5, b=-3</math>
+<math>-1*-16=16</math> <math>\Rightarrow</math> Doesn't work
+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>.
+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>
-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~
+-skibbysiggy
 == See Also ==
 {{AMC12 box|year=2015|ab=A|num-b=17|num-a=19}}

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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