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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
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== Solution ==
 
== Solution ==
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A quadratic equation has no real solutions if and only if the discriminant is nonpositive. Therefore:
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A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>Since <math>x^2+bx+c=0</math> does not have real solutions, we have <math>b^2\leq 4c.</math></li><p>
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  <li>Since <math>x^2+cx+b=0</math> does not have real solutions, we have <math>c^2\leq 4b.</math></li><p>
 +
</ol>
 +
Squaring the first inequality, we get <math>b^4\leq 16c^2.</math> Multiplying the second inequality by <math>16,</math> we get <math>16c^2\leq 64b.</math> Combining these results, we get <cmath>b^4\leq 16c^2\leq 64b.</cmath>
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==See Also==
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{{AMC10 box|year=2021 Fall|ab=A|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 19:08, 22 November 2021

Problem

How many ordered pairs of positive integers $(b,c)$ exist where both $x^2+bx+c=0$ and $x^2+cx+b=0$ do not have distinct, real solutions?

$\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad$

Solution

A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:

  1. Since $x^2+bx+c=0$ does not have real solutions, we have $b^2\leq 4c.$

  2. Since $x^2+cx+b=0$ does not have real solutions, we have $c^2\leq 4b.$

Squaring the first inequality, we get $b^4\leq 16c^2.$ Multiplying the second inequality by $16,$ we get $16c^2\leq 64b.$ Combining these results, we get \[b^4\leq 16c^2\leq 64b.\]

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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

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