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

(added better solution, slight revision to old solution)
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== Solution 2 ==
 
== Solution 2 ==
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One approach is to plug in some roots.
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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>  
+
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>.

Latest revision as of 15:05, 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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