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Difference between revisions of "1965 AHSME Problems/Problem 7"

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\textbf{(E) }\ -\frac{b}{c} </math>
 
\textbf{(E) }\ -\frac{b}{c} </math>
  
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== Solution ==
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== Solution 1 ==
  
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Using Vieta's formulas, we can write the sum of the roots of any quadratic equation in the form <math>ax^2+bx+c = 0</math> as <math>\frac{-b}{a}</math>, and the product as <math>\frac{c}{a}</math>.
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Using [[Vieta's formulas]], we can write the sum of the roots of any quadratic equation in the form <math>ax^2+bx+c = 0</math> as <math>\frac{-b}{a}</math>, and the product as <math>\frac{c}{a}</math>.
  
 
If <math>r</math> and <math>s</math> are the roots, then the sum of the reciprocals of the roots is <math>\frac{1}{r} + \frac{1}{s} = \frac{r+s}{rs}</math>.
 
If <math>r</math> and <math>s</math> are the roots, then the sum of the reciprocals of the roots is <math>\frac{1}{r} + \frac{1}{s} = \frac{r+s}{rs}</math>.
  
 
Applying the formulas, we get <math>\frac{\frac{-b}{a}}{\frac{c}{a}}</math>, or <math>\frac {-b}{c}</math> => <math>\boxed{E}</math>.
 
Applying the formulas, we get <math>\frac{\frac{-b}{a}}{\frac{c}{a}}</math>, or <math>\frac {-b}{c}</math> => <math>\boxed{E}</math>.
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== Solution 2 ==
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Let <math>f(x)=ax^2+bx+c</math>. <math>f(x)</math> has roots <math>r</math> and <math>s</math>, and so <math>x^2f(\frac{1}{x})</math> has roots <math>\frac{1}{r}</math> and <math>\frac{1}{s}</math>. Because <math>x^2f(\frac{1}{x})=cx^2+bx+a</math>, by [[Vieta's formulas]], <math>\frac{1}{r}+\frac{1}{s}=\boxed{\frac{-b}{c}}</math>, which is answer choice <math>\fbox{E}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1965|num-b=6|num-a=8}}
 
{{AHSME 40p box|year=1965|num-b=6|num-a=8}}
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{{MAA Notice}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 16:04, 18 July 2024

Problem

The sum of the reciprocals of the roots of the equation $ax^2 + bx + c = 0$ is:

$\textbf{(A)}\ \frac {1}{a} + \frac {1}{b} \qquad \textbf{(B) }\ - \frac {c}{b} \qquad \textbf{(C) }\ \frac{b}{c}\qquad \textbf{(D) }\ -\frac{a}{b}\qquad \textbf{(E) }\ -\frac{b}{c}$

Solution 1

Using Vieta's formulas, we can write the sum of the roots of any quadratic equation in the form $ax^2+bx+c = 0$ as $\frac{-b}{a}$, and the product as $\frac{c}{a}$.

If $r$ and $s$ are the roots, then the sum of the reciprocals of the roots is $\frac{1}{r} + \frac{1}{s} = \frac{r+s}{rs}$.

Applying the formulas, we get $\frac{\frac{-b}{a}}{\frac{c}{a}}$, or $\frac {-b}{c}$ => $\boxed{E}$.

Solution 2

Let $f(x)=ax^2+bx+c$. $f(x)$ has roots $r$ and $s$, and so $x^2f(\frac{1}{x})$ has roots $\frac{1}{r}$ and $\frac{1}{s}$. Because $x^2f(\frac{1}{x})=cx^2+bx+a$, by Vieta's formulas, $\frac{1}{r}+\frac{1}{s}=\boxed{\frac{-b}{c}}$, which is answer choice $\fbox{E}$.

See Also

1965 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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