Difference between revisions of "1958 AHSME Problems/Problem 41"
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== Problem == | == Problem == | ||
− | The roots of <math> Ax^2 | + | The roots of <math> Ax^2 + Bx + C = 0</math> are <math> r</math> and <math> s</math>. For the roots of |
<math> x^2+px +q =0</math> | <math> x^2+px +q =0</math> | ||
to be <math> r^2</math> and <math> s^2</math>, <math> p</math> must equal: | to be <math> r^2</math> and <math> s^2</math>, <math> p</math> must equal: | ||
− | <math> \textbf{(A)}\ \frac{B^2 | + | <math> \textbf{(A)}\ \frac{B^2 - 4AC}{A^2}\qquad |
− | \textbf{(B)}\ \frac{B^2 | + | \textbf{(B)}\ \frac{B^2 - 2AC}{A^2}\qquad |
− | \textbf{(C)}\ \frac{2AC | + | \textbf{(C)}\ \frac{2AC - B^2}{A^2}\qquad \\ |
− | \textbf{(D)}\ B^2 | + | \textbf{(D)}\ B^2 - 2C\qquad |
− | \textbf{(E)}\ 2C | + | \textbf{(E)}\ 2C - B^2</math> |
− | |||
== Solution == | == Solution == |
Revision as of 23:25, 13 March 2015
Problem
The roots of are
and
. For the roots of
to be and
,
must equal:
Solution
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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