Difference between revisions of "2019 AMC 12A Problems/Problem 14"
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<math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math> | <math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math> | ||
− | by using the quadratic formula on the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of two (not necessarily distinct) factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of <math>4</math> when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = 3 \pm i</math> and <math>|c| = \boxed{\textbf{(E) } \sqrt{10}}</math>. | + | by using the [[quadratic formula]] on each of the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of two (not necessarily distinct) factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of <math>4</math> when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = 3 \pm i</math> and <math>|c| = \boxed{\textbf{(E) } \sqrt{10}}</math>. |
==See Also== | ==See Also== |
Latest revision as of 22:02, 1 January 2020
Problem
For a certain complex number , the polynomial has exactly 4 distinct roots. What is ?
Solution
The polynomial can be factored further broken down into
by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term must be a product of any combination of two (not necessarily distinct) factors from the set: and . We need the two factors to yield a constant term of when multiplied together. The only combinations that work are and , or and . When multiplied together, the polynomial is either or . Therefore, and .
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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All AMC 12 Problems and Solutions |
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