Square roots

**sfspitfire23** · Aug 3rd 2010, 08:06 AM

Solve $\frac{3}{\sqrt{2}} \times (2\sqrt{3}-6\sqrt{3})$

First, I rationalize the first fraction and get $\frac{3\sqrt{2}}{2}$ and then multiply to get $\frac{-9\sqrt{6}}{2}$

Is this correct?

**eumyang** · Aug 3rd 2010, 08:12 AM

$\begin{aligned} \dfrac{3}{\sqrt{2}} \times (2\sqrt{3}-6\sqrt{3}) &= \dfrac{3}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}}(-4\sqrt{3}) \\ &= \dfrac{3\sqrt{2}}{2} \cdot (-4\sqrt{3}) \\ &= \dfrac{12\sqrt{6}}{2} \\ &= -6\sqrt{6} \end{aligned}$

EDIT: Miscopied the problem...

**yeKciM** · Aug 3rd 2010, 08:14 AM

Originally Posted by sfspitfire23

Solve $\frac{3}{\sqrt{2}} \times (2\sqrt{3}-6\sqrt{3})$

First, I rationalize the first fraction and get $\frac{3\sqrt{2}}{2}$ and then multiply to get $\frac{-9\sqrt{6}}{2}$

Is this correct?

u have done this :

$\sqrt{2} \sqrt{3} = \sqrt{6}$

i hope u didn't

result should be :

$-6\sqrt{2}\sqrt{3}$

**sfspitfire23** · Aug 3rd 2010, 08:22 AM

eumyang i think you did the wrong problem

**eumyang** · Aug 3rd 2010, 08:27 AM

Yeah, fixed it now.

Originally Posted by yeKciM

u have done this :

$\sqrt{2} \sqrt{3} = \sqrt{6}$

i hope u didn't

Why not? It's perfectly valid.

result should be :

$-6\sqrt{2}\sqrt{3}$

When I learned this, it's okay to leave it like this: $-6\sqrt{6}$ because there are no perfect square factors of 6.

**yeKciM** · Aug 3rd 2010, 08:31 AM

Originally Posted by eumyang

Yeah, fixed it now.

Why not? It's perfectly valid.

When I learned this, it's okay to leave it like this: $-6\sqrt{6}$ because there are no perfect square factors of 6.

aaaaah sorry I was looking in something else ... sorry (for some reason i was going to addition)

lol here like 38+°C and I'm starting to act very very stupid

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