Difference between revisions of "1965 AHSME Problems/Problem 7"
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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{ | + | Applying the formulas, we get <math>\frac{\frac{-b}{a}}{\frac{c}{a}}</math>, or <math>\frac {-b}{c}</math> => <math>\boxed{E}</math>. |
Revision as of 12:49, 16 July 2024
Problem
The sum of the reciprocals of the roots of the equation 
is:
Solution
Using Vieta's formulas, we can write the sum of the roots of any quadratic equation in the form 
as 
, and the product as 
.
If 
and 
are the roots, then the sum of the reciprocals of the roots is 
.
Applying the formulas, we get 
, or 
=> 
.
