1989 AHSME Problems/Problem 28


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

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


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 $\rm{(D)}$.

Solution 2

$t^2-9t+1=0$: We treat $\tan(x_1)$ and $\tan(x_2)$ as the roots of our equation. Because $\tan(x_1) \times \tan(x_2) = 1$ by Vieta's formula, $x_1 + x_2 = 0.5\pi$. Because the principal values of $x_1$ and $x_2$ are acute and our range for $x$ is $[0,2\pi]$, we have four values of $x$ that satisfy the quadratic: $x_1, x_2, x_1+\pi, x_2+\pi.$ Summing these, we obtain $2(x_1+x_2) + 2\pi$. Using the fact that $x_1+x_2=0.5\pi$, we get $2(0.5\pi) + 2\pi = 3\pi.$

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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