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

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== Solution ==
 
== Solution ==
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The roots of <math>t^2-9t+1=0</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</math> are between <math>0</math> and <math>\tfrac\pi{2}</math>, and the two other roots are <math>\pi+x_1,\ \pi+x_2</math>.
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The roots of <math>t^2-9t+1=0</math> are positive and distinct, so by considering the graph of <math>y=\tan x</math>, the smallest two roots of the original equation <math>x_1,\ x_2</math> are between <math>0</math> and <math>\tfrac\pi{2}</math>, and the two other roots are <math>\pi+x_1, \pi+x_2</math>.
  
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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>.
  
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==Solution 2==
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<math>t^2-9t+1=0</math>
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<math>t^2-9t+1=0</math>:
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We treat <math>tan(x_1)</math> and <math>tan(x_2)</math> as the roots of our equation  
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We treat <math>\tan(x_1)</math> and <math>\tan(x_2)</math> as the roots of our equation.
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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>.
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Because <math>\tan(x_1) \times \tan(x_2) = 1</math> by Vieta's formula, <math>x_1 + x_2 = 0.5\pi</math>.
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Because the principle values of x_1 and x_2 are acute and our range for x is <math>[0,2pi{2}]</math>,  
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Because the principal values of <math>x_1</math> and <math>x_2</math> are acute and our range for <math>x</math> is <math>[0,2\pi]</math>,  
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we have four values of x that satisfy the quadratic:
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we have four values of <math>x</math> that satisfy the quadratic:
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<math>x_1, x_2, x_1+pi{2}, x_2+</math>pi{2}<math>
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<math>x_1, x_2, x_1+\pi, x_2+\pi.</math>
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Summing these, we obtain </math>2(x_1+x_2) + 2pi{2}<math>.
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Summing these, we obtain <math>2(x_1+x_2) + 2\pi</math>.
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Using the fact that </math>x_1+x_2=0.5pi{2}<math>
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Using the fact that <math>x_1+x_2=0.5\pi</math>,
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</math>2(0.5pi{2})<math> + </math>2pi{2}<math> = </math>3pi{2}$
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we get <math>2(0.5\pi) + 2\pi = 3\pi.</math>
  
 
== See also ==
 
== See also ==
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[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}
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tan^2(x) -9tan(x)+1
 
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We treat tan(x1) and tan(x2) as the roots of our equation
 
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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],
 
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we have four values of x that satisfy the quadratic:
 
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x1, x2, x1+pi, x2+pi
 
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Summing these, we obtain 2(x1+x2) + 2pi.
 
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Using the fact that x1+x2=0.5pi
 
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2(0.5pi) + 2pi = 3pi
 

Latest revision as of 20:14, 2 March 2019

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 $\rm{(D)}$.

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 (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

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