+0

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

0
380
1

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
+92174
+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.

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

this is first quad so third quad equivalent is

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

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

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

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

I think that is right

Melody  Feb 26, 2015
Sort:

### 1+0 Answers

#1
+92174
+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.

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

this is first quad so third quad equivalent is

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

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

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

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

I think that is right

Melody  Feb 26, 2015

### 21 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details