Difference between revisions of "1962 AHSME Problems/Problem 11"
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==Solution== | ==Solution== | ||
| − | {{ | + | Call the two roots <math>r</math> and <math>s</math>, with <math>r \ge s</math>. |
| + | By Vieta's formulas, <math>p=r+s</math> and <math>(p^2-1)/4=rs.</math> | ||
| + | (Multiplying both sides of the second equation by 4 gives <math>p^2-1=4rs</math>.) | ||
| + | The value we need to find, then, is <math>r-s</math>. | ||
| + | Since <math>p=r+s</math>, <math>p^2=r^2+2rs+s^2</math>. | ||
| + | Subtracting <math>p^2-1=4rs</math> from both sides gives <math>1=r^2-2rs+s^2</math>. | ||
| + | Taking square roots, <math>r-s=1 \Rightarrow \boxed{\textbf{(B)}}</math>. | ||
Revision as of 12:13, 16 April 2014
Problem
The difference between the larger root and the smaller root of 
is:
Solution
Call the two roots 
and 
, with 
.
By Vieta's formulas, 
and 
(Multiplying both sides of the second equation by 4 gives 
.)
The value we need to find, then, is 
.
Since , 
.
Subtracting from both sides gives 
.
Taking square roots, 
.
