1989 AHSME Problems/Problem 28
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 
.
We treat 
and 
as the roots of our equation
Because * = 
by Vieta's formula, 
.
Because the principle values of x_1 and x_2 are acute and our range for x is ![$[0,2pi{2}]$](http://latex.artofproblemsolving.com/9/f/5/9f5178fe503c4117b97d13f7a51bc5d1dbe821f3.png)
,
we have four values of x that satisfy the quadratic:
pi{2}
2(x_1+x_2) + 2pi{2}
x_1+x_2=0.5pi{2}$$ (Error compiling LaTeX. Unknown error_msg)2(0.5pi{2})
2pi{2}
3pi{2}$
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. 
tan^2(x) -9tan(x)+1 We treat tan(x1) and tan(x2) as the roots of our equation Because tan(x1) * tan(x2) = 1 by Vieta's formula, x1 + x2 = 0.5pi. Because the principle values of x1 and x2 are acute and our range for x is [0,2pi], we have four values of x that satisfy the quadratic: x1, x2, x1+pi, x2+pi Summing these, we obtain 2(x1+x2) + 2pi. Using the fact that x1+x2=0.5pi 2(0.5pi) + 2pi = 3pi
