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please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

Guest Mar 10, 2017
edited by Guest  Mar 10, 2017
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4+0 Answers

 #1
avatar+18599 
0

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

 

\(\sqrt{-2}\cdot \sqrt{-12} = \sqrt{(-2)\cdot(-12)}=\sqrt{24}=\sqrt{4\cdot 6}=\sqrt{4}\cdot \sqrt{6}=2\cdot \sqrt{6}=4.89897948557\)

 

laugh

heureka  Mar 10, 2017
 #3
avatar+26231 
0

Need to be careful here heureka

 

\(\sqrt{-2}\rightarrow i\sqrt2\\ \sqrt{-12}\rightarrow 2i\sqrt3\)

 

Multiply these together and you should get a negative result.

Alan  Mar 10, 2017
 #2
avatar+10614 
+5

....And also , I think

   sqrt(-2) x sqrt(-12) = i sqrt 2 x i sqrt 12  = i^2 sqrt(24) = i^2 (2) sqrt 6 = -2 sq rt 6

ElectricPavlov  Mar 10, 2017
 #4
avatar+18599 
0

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

\(\sqrt{-2}\cdot \sqrt{-12} =\ ?\)

 

\(\sqrt{z} \qquad z \in \mathbb{Q}\) has two solutions in the complex plane.

 

 

1. Wolfram Alpha calculate for \(\sqrt{-2}\):

 

 

2. Wolfram Alpha calculate for \(\sqrt{-12}\):

 

So the product is:

\(\begin{array}{|rcll|} \hline && \sqrt{-2}\cdot \sqrt{-12} \\ &=& (\pm i\cdot\sqrt{2})\times (\pm 2i\sqrt{3}) \\ \hline \end{array} \)

 

Here we have four solutions:

\(\begin{array}{|rcll|} \hline 1. & (+ i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& -2\sqrt{6} \\ 2. & (+ i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& +2\sqrt{6} \\ 3. & (- i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& +2\sqrt{6} \\ 4. & (- i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& -2\sqrt{6} \\ \hline \end{array}\)

 

and finally:

\(\begin{array}{|rcll|} \hline \sqrt{-2}\cdot \sqrt{-12} = \pm 2\sqrt{6} \\ \hline \end{array}\)

 

laugh

heureka  Mar 10, 2017

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