Difference between revisions of "1965 AHSME Problems/Problem 7"
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| + | == Problem == | ||
| + | |||
| + | The sum of the reciprocals of the roots of the equation <math>ax^2 + bx + c = 0</math> is: | ||
| + | |||
| + | <math>\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} </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
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>. | 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>\ | + | Applying the formulas, we get <math>\frac{\frac{-b}{a}}{\frac{c}{a}}</math>, or <math>\frac {-b}{c}</math> => <math>\boxed{E}</math>. |
Latest revision as of 13:49, 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 
=> 
.
