Difference between revisions of "2023 AMC 10B Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
− | <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are 6 positives and 5 negatives. | + | <math>P(x)</math> is a product of <math>(x-r_n)</math> or 10 terms. When <math>x < 1</math>, all terms are <math>< 0</math>, but <math>P(x) > 0</math> because there is an even number of terms. The sign keeps alternating <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are <math>\boxed{\textbf{ 6}}</math> positives and 5 negatives. <math>\boxed{\textbf{(C) 6}}</math> |
~<math>\textbf{Techno}\textcolor{red}{doggo}</math> | ~<math>\textbf{Techno}\textcolor{red}{doggo}</math> |
Revision as of 17:41, 15 November 2023
When the roots of the polynomial
are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is positive?
Solution 1
is a product of or 10 terms. When , all terms are , but because there is an even number of terms. The sign keeps alternating . There are 11 intervals, so there are positives and 5 negatives.
~
Solution
Denote by the interval for and the interval .
Therefore, the number of intervals that is positive is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)