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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 1 (Casework) ==
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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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| − | We apply casework to the value of <math>b:</math>
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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 <math>\boxed{\textbf{(B) } 6}</math> ordered pairs <math>(b,c),</math> namely <math>(1,1),(1,2),(2,1),(2,2),(3,3),</math> and <math>(4,4).</math>
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| − | ~MRENTHUSIASM
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| − | == Solution 2 (Graphing) ==
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| − | Similar to Solution 1, use the discriminant to get <math>b^2\leq 4c</math> and <math>c^2\leq 4b</math>. These can be rearranged to <math>c\geq \frac{1}{4}b^2</math> and <math>b\geq \frac{1}{4}c^2</math>. Now, we can roughly graph these two inequalities, letting one of them be the <math>x</math> axis and the other be <math>y</math>.
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| − | The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs:
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| − | <asy>
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| − | unitsize(2);
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| − | Label f;
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| − | f.p=fontsize(6);
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| − | xaxis("$x$",0,5,Ticks(f, 1.0));
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| − | yaxis("$y$",0,5,Ticks(f, 1.0));
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| − | real f(real x)
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| − | {
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| − | return 0.25x^2;
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| − | }
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| − | real g(real x)
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| − | {
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| − | return 2*sqrt(x);
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| − | }
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| − | dot((1,1));
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| − | dot((2,1));
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| − | dot((1,2));
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| − | dot((2,2));
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| − | dot((3,3));
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| − | dot((4,4));
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| − | draw(graph(f,0,sqrt(20)));
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| − | draw(graph(g,0,5));
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| − | </asy>
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| − | We are looking for lattice points (since <math>b</math> and <math>c</math> are positive integers), of which we can count <math>\boxed{\textbf{(B) } 6}</math>.
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| − | ~aop2014
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| − | ==Solution 3 (Oversimplified but Risky)==
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| − | A quadratic equation <math>Ax^2+Bx+C=0</math> has one real solution if and only if <math>\sqrt{B^2-4AC}=0.</math> Similarly, it has imaginary solutions if and only if <math>\sqrt{B^2-4AC}<0.</math> 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 <math>b</math> and <math>c.</math> 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{\textbf{(B) } 6}.</math>
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| − | ~Arcticturn
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| − | == Solution 4 ==
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| − | We need to solve the following system of inequalities:
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| − | <cmath>
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| − | \[
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| − | \left\{
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| − | \begin{array}{ll}
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| − | b^2 - 4 c \leq 0 \\
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| − | c^2 - 4 b \leq 0
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| − | \end{array}
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| − | \right..
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| − | \]
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| − | </cmath>
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| − | Feasible solutions are in the region formed between two parabolas <math>b^2 - 4 c = 0</math> and <math>c^2 - 4 b = 0</math>.
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| − | Define <math>f \left( b \right) = \frac{b^2}{4}</math> and <math>g \left( b \right) = 2 \sqrt{b}</math>.
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| − | Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
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| − | For <math>b = 1</math>, <math>f \left( b \right) = \frac{1}{4}</math> and <math>g \left( b \right) = 2</math>.
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| − | Hence, the feasible <math>c</math> are 1, 2.
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| − | For <math>b = 2</math>, <math>f \left( b \right) = 1</math> and <math>g \left( b \right) = 2 \sqrt{2}</math>.
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| − | Hence, the feasible <math>c</math> are 1, 2.
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| − | For <math>b = 3</math>, <math>f \left( b \right) = \frac{9}{4}</math> and <math>g \left( b \right) = 2 \sqrt{3}</math>.
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| − | Hence, the feasible <math>c</math> is 3.
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| − | For <math>b = 4</math>, <math>f \left( b \right) = 4</math> and <math>g \left( b \right) = 4</math>.
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| − | Hence, the feasible <math>c</math> is 4.
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| − | For <math>b > 4</math>, <math>f \left( b \right) > g \left( b \right)</math>. Hence, there is no feasible <math>c</math>.
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| − | Putting all cases together, the correct answer is <math>\boxed{\textbf{(B) }6}</math>.
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| − | ~Steven Chen (www.professorchenedu.com)
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| − | ==Video Solution by Mathematical Dexterity==
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| − | https://www.youtube.com/watch?v=EkaKfkQgFbI
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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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