Difference between revisions of "1956 AHSME Problems/Problem 7"

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

Revision as of 12:42, 15 March 2023

Problem

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

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


Solution

Dividing the equation by $a\quad(a\neq0)$ gives: $x^2+\frac{b}{a}x+\frac{c}{a}$. Let $r$ and $s$ be the roots of the equation \[r=\frac{1}{s}\] \[rs=1\]

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

See Also

1956 AHSME (ProblemsAnswer KeyResources)
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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