+0

That is all

0
202
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
#1
+19653
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
+26758
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
+12565
+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
+19653
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