$\displaystyle \sqrt{4x^4+36x^2+81}$

shrug, i should know how to do this lol

help :P

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

I get lost at this point

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- May 16th 2008, 01:05 AMsilencecloaksimplifying radical
$\displaystyle \sqrt{4x^4+36x^2+81}$

shrug, i should know how to do this lol

help :P

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

I get lost at this point - May 16th 2008, 02:39 AMIsomorphism
- May 16th 2008, 02:58 AMsilencecloak
- May 16th 2008, 02:59 AMangel.white
For future reference:

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

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

$\displaystyle =9*6*x$

$\displaystyle =54x$

Another way of doing this:

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

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

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

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

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

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

$\displaystyle =54x$ - May 16th 2008, 03:15 AMIsomorphism
Well I am acquainted with the formula

$\displaystyle a^2 + 2ab + b^2 = (a+b)^2$

So when I see $\displaystyle 4x^4+36x^2+81$,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 $\displaystyle 2x^2$ and $\displaystyle 9$ and check if $\displaystyle 36x^2$ is the expected $\displaystyle 2(2x^2)9$. It is...

So yay! :D