Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"

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If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>.
 
If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>.
  
{{AMC12 box|year=2021 Fall|ab=A|num-a=16|num-b=18}}
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{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:48, 23 November 2021

Problem

For how many ordered pairs $(b,c)$ of positive integers does neither $x^2+bx+c=0$ nor $x^2+cx+b=0$ have two distinct real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad$

Solution

If a quadratic equation does not have two distinct real solutions, then its discriminant must be $\le0$. So, $b^2-4c\le0$ and $c^2-4b\le0$. By inspection, there are $\boxed{\textbf{(B) } 6}$ ordered pairs of positive integers that fulfill these criteria: $(1,1)$, $(1,2)$, $(2,1)$, $(2,2)$, $(3,3)$, and $(4,4)$.

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 12 Problems and Solutions

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