Difference between revisions of "2023 AMC 10B Problems/Problem 12"
Technodoggo (talk | contribs) (→Solution) |
Technodoggo (talk | contribs) (→Solution) |
||
| Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
| − | 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 +,-,+,-,....,+. 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. |
~Technodoggo | ~Technodoggo | ||
Revision as of 17:51, 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
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.
~Technodoggo
