Difference between revisions of "1958 AHSME Problems/Problem 41"
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By Vieta's, <math>r + s = -\frac{B}{A}</math>, <math>rs = \frac{C}{A}</math>, and <math>r^2 + s^2 = -p</math>. Note that <math>(r+s)^2 = r^2 + s^2 + 2rs</math>. | By Vieta's, <math>r + s = -\frac{B}{A}</math>, <math>rs = \frac{C}{A}</math>, and <math>r^2 + s^2 = -p</math>. Note that <math>(r+s)^2 = r^2 + s^2 + 2rs</math>. | ||
− | Therefore, <math>(r + s)^2 - 2rs = r^2 + s^2</math>, or <math>-\left(\frac{B}{A}\right)^2 - | + | Therefore, <math>(r + s)^2 - 2rs = r^2 + s^2</math>, or <math>-\left(\frac{B}{A}\right)^2 - \frac{2C}{A} = -p</math>. |
Simplifying, <math>\frac{B^2 - 2CA}{A^2} = -p</math>. | Simplifying, <math>\frac{B^2 - 2CA}{A^2} = -p</math>. |
Latest revision as of 14:10, 22 February 2018
Problem
The roots of are and . For the roots of
to be and , must equal:
Solution
By Vieta's, , , and . Note that .
Therefore, , or .
Simplifying, .
Finally, multiply both sides by to get , making the answer .
See Also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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All AHSME Problems and Solutions |
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