Difference between revisions of "2007 AMC 12A Problems/Problem 21"
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To disprove the others, note that: | To disprove the others, note that: | ||
− | *<math>\mathrm{B}</math>: then <math>b = \frac {-b}a</math>, which | + | *<math>\mathrm{B}</math>: then <math>b = \frac {-b}a</math>, which is not necessarily true. |
− | *<math>\mathrm{C}</math>: the [[y-intercept]] is <math>c</math>, so <math>c = \frac ca</math> | + | *<math>\mathrm{C}</math>: the [[y-intercept]] is <math>c</math>, so <math>c = \frac ca</math>, not necessarily true. |
*<math>\mathrm{D}</math>: an [[x-intercept]] of the graph is a root of the polynomial, but this excludes the other root. | *<math>\mathrm{D}</math>: an [[x-intercept]] of the graph is a root of the polynomial, but this excludes the other root. | ||
*<math>\mathrm{E}</math>: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2. | *<math>\mathrm{E}</math>: the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2. |
Revision as of 17:16, 2 January 2008
Problem
The sum of the zeros, the product of the zeros, and the sum of the coefficients of the function are equal. Their common value must also be which of the following?
Solution
By Vieta's formulas, the sum of the roots of a quadratic equation is , the product of the zeros is , and the sum of the coefficients is . Setting equal the first two tells us that . Thus, , so the common value is also equal to the coefficient of .
To disprove the others, note that:
- : then , which is not necessarily true.
- : the y-intercept is , so , not necessarily true.
- : an x-intercept of the graph is a root of the polynomial, but this excludes the other root.
- : the mean of the x-intercepts will be the sum of the roots of the quadratic divided by 2.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |