$\displaystyle \sqrt{4x^4+36x^2+81}$
shrug, i should know how to do this lol
help :P
$\displaystyle 2x^2+9\sqrt{36x^2}$
I get lost at this point
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:
$\displaystyle 9*\sqrt{36x^2}$
$\displaystyle =\sqrt{9^2}*\sqrt{36x^2}$
$\displaystyle =\sqrt{81}*\sqrt{36x^2}$
$\displaystyle =\sqrt{36*81*x^2}$
$\displaystyle =\sqrt{2916*x^2}$
$\displaystyle =\sqrt{2916}*\sqrt{x^2}$
$\displaystyle =54x$
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!