Difference between revisions of "1960 AHSME Problems/Problem 31"
Rockmanex3 (talk | contribs) (Solution to Problem 31) |
Rockmanex3 (talk | contribs) m (→Solution) |
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Substitute <math>c</math> and <math>b</math> back to get | Substitute <math>c</math> and <math>b</math> back to get | ||
<cmath>x^4 + 6x^2 + 25</cmath> | <cmath>x^4 + 6x^2 + 25</cmath> | ||
− | Thus, <math>p=6</math> and <math>q=25</math>, so the answer is <math>\boxed{\textbf{(D)}}</math> | + | Thus, <math>p=6</math> and <math>q=25</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>. |
==See Also== | ==See Also== | ||
{{AHSME 40p box|year=1960|num-b=30|num-a=32}} | {{AHSME 40p box|year=1960|num-b=30|num-a=32}} |
Revision as of 18:44, 12 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 | |
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All AHSME Problems and Solutions |