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

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Let <math>r_1</math> and <math>r_2</math> be the roots of the given polynomial. By Vieta's, <math>r_1+r_2=-k</math>, and <math>r_1r_2=36</math>.  
 
Let <math>r_1</math> and <math>r_2</math> be the roots of the given polynomial. By Vieta's, <math>r_1+r_2=-k</math>, and <math>r_1r_2=36</math>.  
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The <math>5</math> positive pairs of <math>(r_1, r_2)</math> that satisfy the second constraint are: <math>(1, 36), (2, 18), (3, 12), (4, 9),</math> and <math>(6, 6).</math>
 
The <math>5</math> positive pairs of <math>(r_1, r_2)</math> that satisfy the second constraint are: <math>(1, 36), (2, 18), (3, 12), (4, 9),</math> and <math>(6, 6).</math>
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We multiply this number by <math>2</math> to account for the negative numbers that will multiply together to give us <math>36</math>, and subtract <math>2</math>, since <math>(6, 6)</math> is both a double root and is counted twice.  
 
We multiply this number by <math>2</math> to account for the negative numbers that will multiply together to give us <math>36</math>, and subtract <math>2</math>, since <math>(6, 6)</math> is both a double root and is counted twice.  
  
 
Therefore, the answer is <math>5\cdot2-2=\boxed{\textbf{(B) }8}</math> pairs.
 
Therefore, the answer is <math>5\cdot2-2=\boxed{\textbf{(B) }8}</math> pairs.
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-Benedict T (countmath1)
 
-Benedict T (countmath1)

Revision as of 23:20, 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

See Also

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

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