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| − | == Problem ==
| + | #REDIRECT [[2021_Fall_AMC_12A_Problems/Problem_17]] |
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| − | How many ordered pairs of positive integers <math>(b,c)</math> exist where both <math>x^2+bx+c=0</math> and <math>x^2+cx+b=0</math> do not have distinct, real solutions?
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| − | <math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 12 \qquad</math>
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| − | == Solution ==
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| − | A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
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| − | <ol style="margin-left: 1.5em;">
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| − | <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>
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| − | </ol>
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| − | 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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| − | Note that:
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| − | * If <math>b=1,</math> then <math>1\leq 16c^2\leq 64,</math> from which <math>c=1,2.</math>
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| − | * If <math>b=2,</math> then <math>16\leq 16c^2\leq 128,</math> from which <math>c=1,2.</math>
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| − | * If <math>b=3,</math> then <math>81\leq 16c^2\leq 192,</math> from which <math>c=3.</math>
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| − | * If <math>b=4,</math> then <math>256\leq 16c^2\leq 256,</math> from which <math>c=4.</math>
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| − | Together, there are
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| − | ~MRENTHUSIASM
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| − | ==Solution 1(Oversimplified but risky)==
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| − | A quadratic equation has one solution if and only if <math>\sqrt {b^2-4ax}</math> is <math>0</math>. Similarly, it is imaginary if and only if <math>\sqrt {b^2-4ax}</math> is less than one. We proceed as following:
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| − | We want both <math>x^2+bx+c</math> to be <math>1</math> value or imaginary and <math>x^2+cx+b</math> to be <math>1</math> value or imaginary. <math>x^2+4x+4</math> is one such case since <math>\sqrt {b^2-4ac}</math> is <math>0</math>. Also, <math>x^2+3x+3, x^2+2x+2, x^2+x+1</math> are always imaginary for both b and c. We also have <math>x^2+x+2</math> along with <math>x^2+2x+1</math> since the latter has one solution, while the first one is imaginary. Therefore, we have <math>6</math> total ordered pairs of integers, which is <math>\boxed {(B) 6}</math>
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| − | ~Arcticturn
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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}}
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