Hello forum, vaironxxrd here.

I somehow con fussed myself trying to solve radicals I got many more questions but I will ask the simple ones first.

Problem:

What I believe is the solution is

Procedure:

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- Oct 12th 2011, 06:56 AMvaironxxrdRadicals
Hello forum, vaironxxrd here.

I somehow con fussed myself trying to solve radicals I got many more questions but I will ask the simple ones first.

Problem:

What I believe is the solution is

Procedure: - Oct 12th 2011, 07:07 AMTheChazRe: Radicals
Use the property that, for a > 0 and b > 0,

√(ab) = √(a) * √(b)

So √(12), which is √(4*3) = √(4) * √(3) = 2√3 - Oct 12th 2011, 07:09 AMearbothRe: Radicals
- Oct 12th 2011, 08:03 AMvaironxxrdRe: Radicals
Thanks for clarifying that.

To make sure I'm understand this here are two problems .

Problem:

Solve:

A more complicated problem for me.

Problem:

I was testing the results and this works, but can you guys explain why should I place the 4 out and aware the 9.

Solution:

- Oct 12th 2011, 02:17 PMHallsofIvyRe: Radicals
- Oct 12th 2011, 04:25 PMvaironxxrdRe: Radicals
- Oct 12th 2011, 06:16 PMQuackyRe: Radicals
- Oct 12th 2011, 07:31 PMTheChazRe: Radicals
- Oct 13th 2011, 01:04 PMHallsofIvyRe: Radicals
What you

**wrote**is actually !!! Put everything you want in the square root in { }: \sqrt{16n^2} gives if n is 0 or positive and -4n if n is negative. More generally where |n| is the "absolute value".

Quote:

and

I don't need the absolute value for y because is never negative. - Oct 13th 2011, 01:57 PMskeeterRe: RadicalsQuote:

I don't need the absolute value for y because is never negative.

I was told by someone much more knowledgeable than myself that you don't need the absolute value for x either, because x must be greater than or equal to zero in the original expression.