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

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~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
 
~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
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==Alternate Solution==
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Note that <math>k</math> must be an integer. By the quadratic formula, <math>x=\frac{-k \pm \sqrt{k^2-144}}{2}.</math> Since <math>144</math> is a multiple of
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<math>4</math>, <math>k</math> and <math>k^2-144</math> have the same parity, so <math>x</math> is an integer if and only if <math>k^2-144</math> is a perfect square.
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Let k^2-144=n^2. Then, <math>(k+n)(k-n)=144. Since </math>k<math> is an integer and </math>144<math> is even, </math>k+n<math> and </math>k-n<math> must both be even. Assuming that </math>k<math> is positive, we get </math>5<math> possible values of </math>k+n<math>, namely </math>2, 4, 8, 6, 12<math>, which will give distinct positive values of </math>k<math>, but </math>k+n=12<math> gives </math>k+n=k-n<math> and </math>n=0<math>, giving </math>2<math> identical integer roots. Therefore, there are </math>4<math> distinct positive values of </math>k.<math> Multiplying that by </math>2<math> to take the negative values into account, we get </math>4*2=\boxed{8}<math> values of </math>k.$
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pianoboy
  
 
== See Also ==
 
== See Also ==

Revision as of 23:20, 17 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

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 ~$\color{magenta} zoomanTV$

Alternate Solution

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$4*2=\boxed{8}$values of$k.$

pianoboy

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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