Difference between revisions of "1957 AHSME Problems/Problem 27"
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<math>\textbf{(A)}\ -\frac{p}{q} \qquad \textbf{(B)}\ \frac{q}{p}\qquad \textbf{(C)}\ \frac{p}{q}\qquad \textbf{(D)}\ -\frac{q}{p}\qquad\textbf{(E)}\ pq </math> | <math>\textbf{(A)}\ -\frac{p}{q} \qquad \textbf{(B)}\ \frac{q}{p}\qquad \textbf{(C)}\ \frac{p}{q}\qquad \textbf{(D)}\ -\frac{q}{p}\qquad\textbf{(E)}\ pq </math> | ||
| − | == Solution == | + | == Solution 1 == |
| + | Let <math>f(x)=x^2+px+q</math>. Then, <math>x^2f(\tfrac1 x)</math> equals <math>qx^2+px+1</math> and has roots which are the reciprocals of those of <math>f(x)</math>. Thus, by [[Vieta's Formulas]], the sum of the roots of <math>x^2f(\tfrac1 x)</math> (and thus the sum of the reciprocated roots of <math>f(x)</math>) is <math>\boxed{\textbf{(A) }\frac{-p}{q}}</math>. | ||
| + | |||
| + | == Solution 2 == | ||
One approach is to plug in some roots. | One approach is to plug in some roots. | ||
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The roots are <math>x=2</math> and <math>x=3</math>. | The roots are <math>x=2</math> and <math>x=3</math>. | ||
| − | The sum of the roots is <math>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}</math>. | + | The sum of the reciprocals of the roots is <math>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}</math>. |
In this case, <math>p</math> and <math>q</math> are <math>-5</math> and <math>6</math>. | In this case, <math>p</math> and <math>q</math> are <math>-5</math> and <math>6</math>. | ||
Revision as of 16:04, 25 July 2024
Contents
Problem
The sum of the reciprocals of the roots of the equation 
is:
Solution 1
Let 
. Then, 
equals 
and has roots which are the reciprocals of those of 
. Thus, by Vieta's Formulas, the sum of the roots of (and thus the sum of the reciprocated roots of ) is 
.
Solution 2
One approach is to plug in some roots.
We have 
The roots are 
and 
.
The sum of the reciprocals of the roots is 
.
In this case, 
and 
are 
and 
.
Thus, the answer is 
.
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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. 
