1989 AHSME Problems/Problem 28

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 $\rm{(D)}$ .

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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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

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1989 AHSME Problems/Problem 28

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