Difference between revisions of "1965 AHSME Problems/Problem 7"
(see also box) |
(→Solution) |
||
| Line 9: | Line 9: | ||
\textbf{(E) }\ -\frac{b}{c} </math> | \textbf{(E) }\ -\frac{b}{c} </math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | 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>\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>. | ||
| + | |||
| + | == Solution 2 == | ||
| + | |||
| + | 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== | ||
Latest revision as of 10:05, 18 July 2024
Contents
Problem
The sum of the reciprocals of the roots of the equation 
is:
Solution 1
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 
=> 
.
Solution 2
Let 
. 
has roots and , and so 
has roots 
and 
. Because 
, by Vieta's formulas, 
, which is answer choice 
.
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 | ||
