Latest revision as of 11:46, 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, $r=\frac{1}{s} \Rightarrow rs=1$ .

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
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
All AHSME Problems and Solutions

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 ==Solution==
-Dividing the equation by <math>a\quad(a\neq0)</math> gives: <math>x^2+\frac{b}{a}x+\frac{c}{a}</math>.
+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>.
-Let <math>r</math> and <math>s</math> be the roots of the equation
-<cmath>r=\frac{1}{s}</cmath> <cmath>rs=1</cmath>
-From [[Vieta's formula]], <math>rs=\frac{c}{a}\Rightarrow\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math>
+Letting <math>r</math> and <math>s</math> be the respective roots to this quadratic, <math>r=\frac{1}{s} \Rightarrow rs=1</math>.
+From [[Vieta's]], <math>rs=\frac{c}{a}</math>, so <math>\frac{c}{a}=1\Rightarrow\boxed{\text{(C) }c=a}</math>
 == See Also==
 {{AHSME box|year=1956|num-b=6|num-a=8}}
 {{MAA Notice}}