Difference between revisions of "1958 AHSME Problems/Problem 41"
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== Solution == | == Solution == | ||
| − | <math>\fbox{}</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 - 2\frac{C}{A} = -p</math>. | ||
| + | |||
| + | Simplifying, <math>\frac{B^2 - 2CA}{A^2} = -p</math>. | ||
| + | |||
| + | Finally, multiply both sides by <math>p = \frac{2CA - B^2}{A^2}</math>, making the answer <math>\fbox{C}</math>. | ||
== See Also == | == See Also == | ||
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 
, making the answer 
.
See Also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 40 |
Followed by Problem 42 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
