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

Guest Feb 26, 2015

Best Answer 

 #1
avatar+91479 
+5

$$\\tan^2x-9tanx+1=0\\\\
$let y=tanx$\\\\
y^2-9y+1=0\\\\
y=\frac{9\pm\sqrt{81-4}}{2}\\\\
y=\frac{9\pm\sqrt{77}}{2}\\\\
tan(x)=\frac{9+\sqrt{77}}{2}\qquad or \qquad tan(x)=\frac{9-\sqrt{77}}{2}$$

 

x is can have he following values.   They are the roots in radians.

 

    $$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{1.458\: \!749\: \!780\: \!644\: \!212\: \!5}}$$

this is first quad so third quad equivalent is  

 

$${\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.458\: \!749\: \!780\: \!644\: \!212\: \!5}} = {\mathtt{4.600\: \!342\: \!434\: \!234\: \!005\: \!7}}$$

 

$$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{0.112\: \!046\: \!546\: \!150\: \!684\: \!1}}$$

this is first quad so third quad equivalent is  

 

$${\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.112\: \!046\: \!546\: \!150\: \!684\: \!1}} = {\mathtt{3.253\: \!639\: \!199\: \!740\: \!477\: \!3}}$$

 

Now I think these four are all the roots between 0 and 2pi radians so the sum of them would be

 

$$\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{\pi}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)}\right) = {\mathtt{6.505\: \!598\: \!527\: \!339\: \!575\: \!9}}$$

 

 I think that is right  

Melody  Feb 26, 2015
Sort: 

1+0 Answers

 #1
avatar+91479 
+5
Best Answer

$$\\tan^2x-9tanx+1=0\\\\
$let y=tanx$\\\\
y^2-9y+1=0\\\\
y=\frac{9\pm\sqrt{81-4}}{2}\\\\
y=\frac{9\pm\sqrt{77}}{2}\\\\
tan(x)=\frac{9+\sqrt{77}}{2}\qquad or \qquad tan(x)=\frac{9-\sqrt{77}}{2}$$

 

x is can have he following values.   They are the roots in radians.

 

    $$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{1.458\: \!749\: \!780\: \!644\: \!212\: \!5}}$$

this is first quad so third quad equivalent is  

 

$${\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1.458\: \!749\: \!780\: \!644\: \!212\: \!5}} = {\mathtt{4.600\: \!342\: \!434\: \!234\: \!005\: \!7}}$$

 

$$\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)} = {\mathtt{0.112\: \!046\: \!546\: \!150\: \!684\: \!1}}$$

this is first quad so third quad equivalent is  

 

$${\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.112\: \!046\: \!546\: \!150\: \!684\: \!1}} = {\mathtt{3.253\: \!639\: \!199\: \!740\: \!477\: \!3}}$$

 

Now I think these four are all the roots between 0 and 2pi radians so the sum of them would be

 

$$\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{\pi}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,\times\,}}\left(\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{\left({\mathtt{9}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{77}}}}\right)}{{\mathtt{2}}}}\right)}\right) = {\mathtt{6.505\: \!598\: \!527\: \!339\: \!575\: \!9}}$$

 

 I think that is right  

Melody  Feb 26, 2015

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