Difference between revisions of "1989 AHSME Problems/Problem 28"

(Solution)
(Solution)
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<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, x_1 + x_2 = 0.5pi{2}.
+
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 the principle values of x1 and x2 are acute and our range for x is [<math>0</math>,<math>2pi{2}</math>],  
+
Because the principle values of x1 and x2 are acute and our range for x is <math>[0,2pi]</math>,  
 
we have four values of x that satisfy the quadratic:
 
we have four values of x that satisfy the quadratic:
x_1, x_2, x_1+<math>pi{2}</math>, x_2+<math>pi{2}</math>
+
<math>x_1, x_2, x_1+pi, x_2+</math>pi<math>
Summing these, we obtain <math>2(x_1+x_2) + 2pi</math>.
+
Summing these, we obtain </math>2(x_1+x_2) + 2pi<math>.
Using the fact that x_1+x_2=<math>0.5pi{2}</math>
+
Using the fact that x_1+x_2=</math>0.5pi<math>
<math>2(0.5pi{2}</math>) + <math>2pi{2}</math> = <math>3pi</math>
+
</math>2(0.5pi)<math> + </math>2pi<math> = </math>3pi$
  
 
== See also ==
 
== See also ==

Revision as of 07:29, 27 June 2016

Problem

Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.

$\mathrm{(A)  \frac{\pi}{2} } \qquad \mathrm{(B) \pi } \qquad \mathrm{(C) \frac{3\pi}{2} } \qquad \mathrm{(D) 3\pi } \qquad \mathrm{(E) 4\pi }$

Solution

The roots of $t^2-9t+1=0$ are positive and distinct, so by considering the graph of $y=\tan x$, the smallest two roots of the original equation $x_1,\ x_2$ are between $0$ and $\tfrac\pi{2}$, and the two other roots are $\pi+x_1,\ \pi+x_2$.

Then from the quadratic equation we discover that the product $\tan x_1\tan x_2=1$ which implies that $\tan(x_1+x_2)$ does not exist. The bounds then imply that $x_1+x_2=\tfrac\pi{2}$. Thus $x_1+x_2+\pi+x_1+\pi+x_2=3\pi$ which is $\rm{(D)}$.

$t^2-9t+1=0$ We treat $tan(x_1)$ and $tan(x_2)$ as the roots of our equation Because $tan(x_1)$ * $tan(x_2)$ = $1$ by Vieta's formula, $x_1 + x_2 = 0.5pi$. 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: $x_1, x_2, x_1+pi, x_2+$pi$Summing these, we obtain$2(x_1+x_2) + 2pi$. Using the fact that x_1+x_2=$0.5pi$$ (Error compiling LaTeX. Unknown error_msg)2(0.5pi)$+$2pi$=$3pi$

See also

1989 AHSME (ProblemsAnswer KeyResources)
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

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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