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-2i(4 - 3i) + 6i

Guest Jun 7, 2017

Best Answer 

 #2
avatar+93627 
+2

Thanks x^2 you are absolutely correct :)

But sometimes the letter i is used as the basic unit of the comples (imaginary) number system.

 

If i is the imaginary number then  \(i=\sqrt{-1}\)

 

Then 

 

\(6i^2-2i\\ =6*-1-2i\\ =-6-2i\)

Melody  Jun 7, 2017
 #1
avatar+2248 
+2

SImplifying this should be relatively simple. 

 

\(-2i(4-3i)+6i\) Distribute the -2i into the parentheses
\(-8i+6i^2+6i\) Combine the like terms
\(-2i+6i^2\) Rearrange the terms so the term with the highest degree is first
\(6i^2-2i\) You are done!
   
   
   
   
   
   
TheXSquaredFactor  Jun 7, 2017
 #2
avatar+93627 
+2
Best Answer

Thanks x^2 you are absolutely correct :)

But sometimes the letter i is used as the basic unit of the comples (imaginary) number system.

 

If i is the imaginary number then  \(i=\sqrt{-1}\)

 

Then 

 

\(6i^2-2i\\ =6*-1-2i\\ =-6-2i\)

Melody  Jun 7, 2017

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