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Difference between revisions of "1957 AHSME Problems/Problem 27"

(Created page with "One approach is to plug in some roots. You have <math>x^{2}-5x+6=0</math> The roots are <math>x=2</math> and <math>x=3</math>. The sum of the roots is <math>\frac{1}{2}+\f...")
 
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One approach is to plug in some roots.
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== Problem ==
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The sum of the reciprocals of the roots of the equation <math>x^2 + px + q = 0</math> is:
  
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You have <math>x^{2}-5x+6=0</math>  
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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>
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== Solution 1 ==
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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>.
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== Solution 2 ==
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One approach is to plug in some values for <math>p</math> and <math>q</math>.
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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>.
  
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The sum of the roots is <math>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}</math>.
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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>.
  
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From there, you can easily tell that the answer is <math>(A)</math>
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Thus, the answer is <math>\boxed{\textbf{(A) }\frac{-p}q}</math>.
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== See Also ==
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{{AHSME 50p box|year=1957|num-b=26|num-a=28}}
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{{MAA Notice}}
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[[Category:AHSME]][[Category:AHSME Problems]]

Latest revision as of 16:05, 25 July 2024

Problem

The sum of the reciprocals of the roots of the equation $x^2 + px + q = 0$ is:

$\textbf{(A)}\ -\frac{p}{q} \qquad \textbf{(B)}\ \frac{q}{p}\qquad \textbf{(C)}\ \frac{p}{q}\qquad \textbf{(D)}\ -\frac{q}{p}\qquad\textbf{(E)}\ pq$

Solution 1

Let $f(x)=x^2+px+q$. Then, $x^2f(\tfrac1 x)$ equals $qx^2+px+1$ and has roots which are the reciprocals of those of $f(x)$. Thus, by Vieta's Formulas, the sum of the roots of $x^2f(\tfrac1 x)$ (and thus the sum of the reciprocated roots of $f(x)$) is $\boxed{\textbf{(A) }\frac{-p}{q}}$.

Solution 2

One approach is to plug in some values for $p$ and $q$.

We have $x^2-5x+6=0$.

The roots are $x=2$ and $x=3$.

The sum of the reciprocals of the roots is $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$.

In this case, $p$ and $q$ are $-5$ and $6$.

Thus, the answer is $\boxed{\textbf{(A) }\frac{-p}q}$.

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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