1962 AHSME Problems/Problem 11: Difference between revisions
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==Solution== | ==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>. | |||
(Another solution is to use the quadratic formula and see that the | |||
roots are <math>\frac{p\pm 1}2</math>, and their difference is 1.) | |||
==See Also== | |||
{{AHSME 40p box|year=1962|before=Problem 10|num-a=12}} | |||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 21:15, 3 October 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, 
.
(Another solution is to use the quadratic formula and see that the
roots are 
, and their difference is 1.)
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
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