Difference between revisions of "1957 AHSME Problems/Problem 27"
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− | + | == Problem == | |
+ | |||
+ | The sum of the reciprocals of the roots of the equation <math>x^2 + px + q = 0</math> is: | ||
− | + | <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 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 values for <math>p</math> and <math>q</math>. | ||
+ | |||
+ | We have <math>x^2-5x+6=0</math>. | ||
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>. | ||
− | + | Thus, the answer is <math>\boxed{\textbf{(A) }\frac{-p}q}</math>. | |
+ | |||
+ | == See Also == | ||
+ | {{AHSME 50p box|year=1957|num-b=26|num-a=28}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:AHSME]][[Category:AHSME Problems]] |
Latest revision as of 16:05, 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 values for and
.
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 | |
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All AHSME Problems and Solutions |
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