• Oct 12th 2011, 06:56 AM
vaironxxrd
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: $\sqrt12$
What I believe is the solution is $3\sqrt2$

Procedure: $12/3=4 , 4/2= 2$
• Oct 12th 2011, 07:07 AM
TheChaz
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 AM
earboth
Quote:

Originally Posted by vaironxxrd
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: $\sqrt12$
What I believe is the solution is $3\sqrt2$

Procedure: $12/3=4 , 4/2= 2$

Your method is definitely wrong. (Square the original number an your result and you'll get 12 = 18 which is very seldom).

$\sqrt{12} = \sqrt{2 \cdot 2 \cdot 3} = \sqrt{2^2} \cdot \sqrt{3} = 2\sqrt{3}$
• Oct 12th 2011, 08:03 AM
vaironxxrd
Quote:

Originally Posted by TheChaz
Use the property that, for a > 0 and b > 0,
√(ab) = √(a) * √(b)

So √(12), which is √(4*3) = √(4) * √(3) = 2√3

Quote:

Originally Posted by earboth
Your method is definitely wrong. (Square the original number an your result and you'll get 12 = 18 which is very seldom).

$\sqrt{12} = \sqrt{2 \cdot * 2 \cdot 3} = \sqrt{2^2} \cdot \sqrt{3} = 2\sqrt{3}$

Thanks for clarifying that.
To make sure I'm understand this here are two problems .

Problem: $\sqrt50$

Solve: $\sqrt{25 \cdot 2 }= 5\sqrt2$

A more complicated problem for me.

Problem: $\sqr48=$
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: $\sqrt{3 \cdot 3 \cdot 4[$
$4\sqrt3$
• Oct 12th 2011, 02:17 PM
HallsofIvy
Quote:

Originally Posted by vaironxxrd
Thanks for clarifying that.
To make sure I'm understand this here are two problems .

Problem: $\sqrt50$

Solve: $\sqrt{25 \cdot 2 }= 5\sqrt2$

A more complicated problem for me.

Problem: $\sqr48=$
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: $\sqrt{3 \cdot 3 \cdot 4[$
$4\sqrt3$

Your problem here is that 3(3)(4)= 4(9)= 36, not 48! 48= 3(16) so $\sqrt{48}= \sqrt{3(16)}= \sqrt{3}\sqrt{16}= 4\sqrt{3}$
• Oct 12th 2011, 04:25 PM
vaironxxrd
Quote:

Originally Posted by HallsofIvy
Your problem here is that 3(3)(4)= 4(9)= 36, not 48! 48= 3(16) so $\sqrt{48}= \sqrt{3(16)}= \sqrt{3}\sqrt{16}= 4\sqrt{3}$

That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
What when working with variables ? Like...

$\sqrt{16n^2}$

and
$\sqrt{81x^3y^4}$
• Oct 12th 2011, 06:16 PM
Quacky
Quote:

Originally Posted by vaironxxrd
That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
What when working with variables ? Like...

$\sqrt16n^2$

and
$\sqrt81x^3y^4$

Why would it be any different?

$\sqrt{16n^2}$
$=\sqrt{4\times 4\times n\times n}$
$=\sqrt{(4n)^2}$
$=4n$
• Oct 12th 2011, 07:31 PM
TheChaz
Quote:

Originally Posted by Quacky
Why would it be any different?

$\sqrt{16n^2}$
$=\sqrt{4\times 4\times n\times n}$
$=\sqrt{(4n)^2}$
$=4n$

.... or 4|n|
• Oct 13th 2011, 01:04 PM
HallsofIvy
Quote:

Originally Posted by vaironxxrd
That is clear, I figure it out after watching a couple of videos but thanks this explanation is even better.
What when working with variables ? Like...

$\sqrt16n^2$

What you wrote is actually $\sqrt{1}(6n^2)= 6n^2$!!! Put everything you want in the square root in { }: \sqrt{16n^2} gives $\sqrt{16n^2}= 4n$ if n is 0 or positive and -4n if n is negative. More generally $\sqrt{16n^2}= 4|n|$ where |n| is the "absolute value".

Quote:

and
$\sqrt81x^3y^4$
$\sqrt{81x^3y^4}= \sqrt{9^2x^2(y^2)^2x}= 9|x|y^2\sqrt{x}$
I don't need the absolute value for y because $|y^2|$ is never negative.
• Oct 13th 2011, 01:57 PM
skeeter
$\sqrt{81x^3y^4}= \sqrt{9^2x^2(y^2)^2x}= 9|x|y^2\sqrt{x}$
I don't need the absolute value for y because $|y^2|$ is never negative.
$\sqrt{81x^3y^4} = 9xy^2\sqrt{x}$