# Thread: what is the next step?

1. ## what is the next step?

square root of 6 (11 square root of 2 minus 2 square root of 2)

I did everything in the paranthesis first

(11 square root of 2 minus 2 square root of 2)

(9 square root of 2)

(which breaks down to 3 square root of 2)

so far I got square root of 6(3 square root of 2)

what do I do next?

2. 1. $\displaystyle \sqrt{6} \left( 11\sqrt{2} - 2\sqrt{2} \right)$
$\displaystyle \sqrt{6} \left( 9\sqrt{2} \right)$
$\displaystyle 9\sqrt{6 \cdot 2} = 9\sqrt{3 \cdot 2 \cdot 2} = 18\sqrt{3}$

How did it break down to $\displaystyle 3\sqrt{2}$ ???

3. Originally Posted by Chop Suey
1. $\displaystyle \sqrt{6} \left( 11\sqrt{2} - 2\sqrt{2} \right)$
$\displaystyle \sqrt{6} \left( 9\sqrt{2} \right)$
$\displaystyle 9\sqrt{6 \cdot 2} = 9\sqrt{3 \cdot 2 \cdot 2} = 18\sqrt{3}$

How did it break down to $\displaystyle 3\sqrt{2}$ ???
How did you get 9 square root of 3 times 2 times 2??????

4. $\displaystyle \sqrt[n]{ab} = \sqrt[n]{a} \cdot \sqrt[n]{b}$

Example:
$\displaystyle \sqrt{2} \cdot \sqrt{6} = \sqrt{6 \cdot 2} = \sqrt{12}$
$\displaystyle \sqrt{3} \cdot \sqrt{9} = \sqrt{3 \cdot 9} = \sqrt{27}$

But:

$\displaystyle \sqrt[3]{3} \cdot \sqrt{2} \neq \sqrt{2 \cdot 3}\ \text{or}\ \sqrt[3]{2 \cdot 3}$

5. Originally Posted by Laura901
square root of 6 (11 square root of 2 minus 2 square root of 2)

I did everything in the paranthesis first

(11 square root of 2 minus 2 square root of 2)

(9 square root of 2)

(which breaks down to 3 square root of 2)

so far I got square root of 6(3 square root of 2)

what do I do next?
take note that 9 is not inside the square root.. so, you should not take the root of 9..