Difference between revisions of "2019 AMC 12A Problems/Problem 14"

Latest revision as of 22:53, 25 July 2022

Problem

For a certain complex number $c$ , the polynomial $P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)$ has exactly 4 distinct roots. What is $|c|$ ?

$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

Solution

The polynomial can be factored further broken down into

$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$

by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$ . We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$ , or $(x - [1+i])$ and $(x - [2-2i])$ . When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$ . Therefore, $c = 3 \pm i$ and $|c| = \boxed{\textbf{(E) } \sqrt{10}}$ .

2019 AMC 12A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC12 box|year=2019|ab=A|num-b=13|num-a=15}}
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Difference between revisions of "2019 AMC 12A Problems/Problem 14"

Latest revision as of 22:53, 25 July 2022

Problem

Solution

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