Difference between revisions of "1960 AHSME Problems/Problem 31"
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Latest revision as of 18:15, 17 May 2018
Problem
For to be a factor of , the values of and must be, respectively:
Solution
Let the other quadratic be , where . Multiply the two quadratics to get Since have no term and no term, the coefficients of these terms must be zero. Substitute and back to get Thus, and , so the answer is .
See Also
1960 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |