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

Revision as of 14:59, 25 July 2024

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

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

One approach is to plug in some roots.

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

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

The sum 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 $\boxed{\textbf{(A) }\frac{-p}q}$ .

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

@@ Line 1: / Line 1: @@
+== 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 ==
 One approach is to plug in some roots.
-You 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>.
@@ Line 9: / Line 16: @@
 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]]