Difference between revisions of "1989 AHSME Problems/Problem 28"
(→Solution) |
(→Solution) |
||
| Line 9: | Line 9: | ||
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>. | 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>. | ||
| + | |||
<math>t^2-9t+1=0</math> | <math>t^2-9t+1=0</math> | ||
We treat <math>tan(x_1)</math> and <math>tan(x_2)</math> as the roots of our equation | We treat <math>tan(x_1)</math> and <math>tan(x_2)</math> as the roots of our equation | ||
| − | Because <math>tan(x_1)</math> * <math>tan(x_2)</math> = <math>1</math> by Vieta's formula, <math>x_1 + x_2 = 0.5pi</math>. | + | Because <math>tan(x_1)</math> * <math>tan(x_2)</math> = <math>1</math> by Vieta's formula, <math>x_1 + x_2 = 0.5pi{2}</math>. |
| − | Because the principle values of | + | Because the principle values of x_1 and x_2 are acute and our range for x is <math>[0,2pi{2}]</math>, |
we have four values of x that satisfy the quadratic: | we have four values of x that satisfy the quadratic: | ||
| − | <math>x_1, x_2, x_1+pi, x_2+</math>pi<math> | + | <math>x_1, x_2, x_1+pi{2}, x_2+</math>pi{2}<math> |
| − | Summing these, we obtain </math>2(x_1+x_2) + 2pi<math>. | + | Summing these, we obtain </math>2(x_1+x_2) + 2pi{2}<math>. |
| − | Using the fact that x_1+x_2= | + | Using the fact that </math>x_1+x_2=0.5pi{2}<math> |
| − | </math>2(0.5pi)<math> + </math>2pi<math> = </math>3pi$ | + | </math>2(0.5pi{2})<math> + </math>2pi{2}<math> = </math>3pi{2}$ |
== See also == | == See also == | ||
Revision as of 06:31, 27 June 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 
.
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
