Revision as of 11:45, 15 March 2023

Problem

The roots of the equation $ax^2 + bx + c = 0$ will be reciprocal if:

$\textbf{(A)}\ a = b \qquad\textbf{(B)}\ a = bc \qquad\textbf{(C)}\ c = a \qquad\textbf{(D)}\ c = b \qquad\textbf{(E)}\ c = ab$

Dividing both sides of the equation by $a\quad(a\neq0)$ gives $x^2+\frac{b}{a}x+\frac{c}{a}$ .

Letting $r$ and $s$ be the respective roots to this quadratic, $\frac{1}{s}\Rightarrowrs=1$ (Error compiling LaTeX. Unknown error_msg).

From Vieta's, $rs=\frac{c}{a}$ , so $\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}$

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AHSME Problems and Solutions

@@ Line 10: / Line 10: @@
 Dividing both sides of the equation by <math>a\quad(a\neq0)</math> gives <math>x^2+\frac{b}{a}x+\frac{c}{a}</math>.
-Letting <math>r</math> and <math>s</math> be the respective roots to this quadratic, <math>r=\frac{1}{s}\Rightarrowrs=1</math>.
+Letting <math>r</math> and <math>s</math> be the respective roots to this quadratic, <math>\frac{1}{s}\Rightarrowrs=1</math>.
 From [[Vieta's]], <math>rs=\frac{c}{a}</math>, so <math>\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math>