Difference between revisions of "1989 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
| + | The roots of <math>t^2-9t+1=0</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</math> are between <math>0</math> and <math>\tfrac\pi{2}</math>, and the two other roots are <math>\pi+x_1,\ \pi+x_2</math>. | ||
| + | Then from the quadratic equation we discover that the product <math>\tan x_1\tan x_2=1</math> which implies that <math>\tan(x_1+x_2)</math> does not exist. The bounds then imply that <math>x_1+x_2=\tfrac\pi{2}</math>. Thus <math>x_1+x_2+\pi+x_1+\pi+x_2=3\pi</math> which is <math>\rm{(D)}</math>. | ||
== See also == | == See also == | ||
{{AHSME box|year=1989|num-b=27|num-a=29}} | {{AHSME box|year=1989|num-b=27|num-a=29}} | ||
Revision as of 02:32, 4 February 2016
Problem
Find the sum of the roots of 
that are between 
and 
radians.
Solution
The roots of 
are positive and distinct, so by considering the graph of 
, the smallest two roots of the original equation 
are between 
and 
, and the two other roots are 
.
Then from the quadratic equation we discover that the product 
which implies that 
does not exist. The bounds then imply that 
. Thus 
which is 
.
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
