• May 16th 2008, 01:05 AM
silencecloak
$\sqrt{4x^4+36x^2+81}$

shrug, i should know how to do this lol

help :P

$2x^2+9\sqrt{36x^2}$

I get lost at this point
• May 16th 2008, 02:39 AM
Isomorphism
Quote:

Originally Posted by silencecloak
$\sqrt{4x^4+36x^2+81}$

shrug, i should know how to do this lol

help :P

$2x^2+9\sqrt{36x^2}$

I get lost at this point

$\sqrt{4x^4+36x^2+81} = \sqrt{(2x^2+9)^2} = 2x^2+9$
• May 16th 2008, 02:58 AM
silencecloak
Quote:

Originally Posted by Isomorphism
$\sqrt{4x^4+36x^2+81} = \sqrt{(2x^2+9)^2} = 2x^2+9$

can you explain to me your thought process as you look at and go through this radical
• May 16th 2008, 02:59 AM
angel.white
For future reference:
$9*\sqrt{36x^2}$

$=9*\sqrt{36}*\sqrt{x^2}$

$=9*6*x$

$=54x$

Another way of doing this:
$9*\sqrt{36x^2}$

$=\sqrt{9^2}*\sqrt{36x^2}$

$=\sqrt{81}*\sqrt{36x^2}$

$=\sqrt{36*81*x^2}$

$=\sqrt{2916*x^2}$

$=\sqrt{2916}*\sqrt{x^2}$

$=54x$
• May 16th 2008, 03:15 AM
Isomorphism
Quote:

Originally Posted by silencecloak
can you explain to me your thought process as you look at and go through this radical

Well I am acquainted with the formula
$a^2 + 2ab + b^2 = (a+b)^2$

So when I see $4x^4+36x^2+81$,I ask myself if 4x^4 is a square of something and if 81 is a square of something. The answer is yes!

So it take $2x^2$ and $9$ and check if $36x^2$ is the expected $2(2x^2)9$. It is...

So yay! :D