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Find all solutions of the equation and express them in the form a + bi. (Enter your answers as a comma-separated list. Simplify your answer completely.) t+1+1/t=0 t=

Guest May 22, 2017
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2+0 Answers

 #1
avatar+1493 
+1

This is your answer that you have been seeking:

\(t=\frac{-1\pm i\sqrt{3}}{2}\)

 

This is the given equation:

 

\(t+1+\frac{1}{t}=0 \)

 

First, let's exclude any extraneous solution. If t=0, then that is an extraneous solution since when you plug t back into the equation, you will divide by 0, which algebra does not allow. Other than that, let's proceed as normal. To get rid of that ugly fraction, let's multiply both sides by t:
 

\(t+1+\frac{1}{t}=0 \)     Multiply by t on both sides

\(t^2+t+1=0\)     This trinomial cannot be factored, so we will have to rely on a different method. I think I'll use the quadratic formula this time:

 

\(a=1,b=1,c=1\)

\(t= {-1 \pm \sqrt{1^2-4(1)(1)} \over 2(1)}\) Simplify from here

\(t={-1\pm\sqrt{-3}\over2}\)Now, let's simplify the square root of -3

\(\sqrt{-3}=\sqrt{(-1)(3)}=i\sqrt{3}\)

 

The solutions are not equal to the extraneous solutions, so these are valid solutions. Therefore the solutions are:
 

\(t = {-1 \pm i\sqrt{3} \over 2}\)

TheXSquaredFactor  May 22, 2017
 #2
avatar+7155 
0

Find all solutions of the equation and express them in the form a + bi. (Enter your answers as a comma-separated list. Simplify your answer completely.) t+1+1/t=0 t=

 

\(t+1+\frac{1}{t}=0\) 

 

\(t^2+t+1=0\\t=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\)

\(t=-\frac{1}{2}\pm\sqrt{\frac{1}{4}-1}\)

\(t=-\frac{1}{2}\pm\sqrt{-\frac{3}{4}}\\t=-\frac{1}{2}\pm \frac{1}{2}\sqrt{-3}\\t=-\frac{1}{2}\pm \frac{1}{2}i\sqrt{3}\)

\(t=\frac{1}{2}(-1\pm i\sqrt{3})\)

 

\(t\in \left\{-\frac{1}{2}+\frac{1}{2}\sqrt{3}\ i\ ,-\frac{1}{2}-\frac{1}{2}\sqrt{3}\ i\right\} \)

 

laugh  !

asinus  May 22, 2017
edited by asinus  May 22, 2017

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