Difference between revisions of "2022 AMC 10B Problems/Problem 7"

Revision as of 21:09, 19 November 2022

The following problem is from both the 2022 AMC 10B #7 and 2022 AMC 12B #4, so both problems redirect to this page.

Problem

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$

Solution 1

Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formulas, we have $p+q=-k$ and $pq=36.$

This shows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are $\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.$ Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$

~stevens0209 ~MRENTHUSIASM ~ $\color{magenta} zoomanTV$

Solution 2

Note that $k$ must be an integer. By the quadratic formula, $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $144$ is a multiple of $4$ , $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.

Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$ , namely $2, 4, 8, 6, 12$ , which will give distinct positive values of $k$ , but $k+n=12$ gives $k+n=k-n$ and $n=0$ , giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4*2=\boxed{\textbf{(B) }8}$ values of $k.$

pianoboy

Solution 3 (Pythagorean Triples)

Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than 0 to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle where k is the hypotenuse, and 12 is a leg. There are four ways we could have this: a $9-12-15$ triangle, a $12-16-20$ triangle, a $5-12-13$ triangle, and a $12-35-37$ triangle. Multiply by two to account for negative k values (since k is being squared) and our answer is $\boxed{\textbf{(B) }8}$ .

Solution 4 (Similar to Solution 1)

Let $r_1$ and $r_2$ be the roots of the given polynomial. By Vieta's, $r_1+r_2=-k$ , and $r_1r_2=36$ .

The $5$ positive pairs of $(r_1, r_2)$ that satisfy the second constraint are: $(1, 36), (2, 18), (3, 12), (4, 9),$ and $(6, 6).$

We multiply this number by $2$ to account for the negative numbers that will multiply together to give us $36$ , and subtract $2$ , since $(6, 6)$ is both a double root and is counted twice.

Therefore, the answer is $5\cdot2-2=\boxed{\textbf{(B) }8}$ pairs.

-Benedict T (countmath1)

Video Solution 1

https://youtu.be/WX871JJbdY4

~Education, the Study of Everything

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 10 Problems and Solutions

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 39: / Line 39: @@
 Therefore, the answer is <math>5\cdot2-2=\boxed{\textbf{(B) }8}</math> pairs.
+-Benedict T (countmath1)
 ==Video Solution 1==

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