Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
m (fixed inaccuracy)
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
One approach is to plug in some roots.
+
== Problem ==
 +
 +
The sum of the reciprocals of the roots of the equation <math>x^2 + px + q = 0</math> is:
  
−
You have <math>x^{2}-5x+6=0</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 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>.
  
−
From there, you can easily tell that the answer is <math>(A)</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 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
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1957_AHSME_Problems/Problem_27&oldid=223648"