+0

# That is all

0
98
4

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
Sort:

### 4+0 Answers

#1
+18599
0

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

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

heureka  Mar 10, 2017
#3
+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
+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
+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:

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

heureka  Mar 10, 2017

### 28 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