Difference between revisions of "2023 AMC 10B Problems/Problem 12"
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are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive? | are removed from the number line, what remains is the union of 11 disjoint open intervals. On how many of these intervals is <math>P(x)</math> positive? | ||
− | ==Solution== | + | ==Solution 1== |
The interval of the alternating signs. <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 they are even number of terms. The sign keep alternates <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are 6 positives and 5 negatives. | The interval of the alternating signs. <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 they are even number of terms. The sign keep alternates <math>+,-,+,-,....,+</math>. There are 11 intervals, so there are 6 positives and 5 negatives. |
Revision as of 18:39, 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
The interval of the alternating signs. is a product of
or 10 terms. When
, all terms are
, but
because they are even number of terms. The sign keep alternates
. There are 11 intervals, so there are 6 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)