Difference between revisions of "1962 AHSME Problems/Problem 11"
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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 22:15, 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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.