Thread: Square roots

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

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

Is this correct?

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

EDIT: Miscopied the problem...

Originally Posted by sfspitfire23

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

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

Is this correct?

u have done this :

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

i hope u didn't


result should be :

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

eumyang i think you did the wrong problem

Yeah, fixed it now.

Originally Posted by yeKciM

u have done this :

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

i hope u didn't

Why not? It's perfectly valid.

result should be :

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

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

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: $\displaystyle -6\sqrt{6}$ 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