An old mathematics book found on Google asserts that:
Square root of $9a + 36\sqrt{3ax} + 108x$
Equals to $3\sqrt{a} + 6\sqrt{3x}$
I find this incorrect, but am not 100% sure. My derivation is:
Let $m = \sqrt{a}$ and $n = 2\sqrt{3x}$, then
$
\begin{align}
9a + 36\sqrt{3ax} + 108x & = 3\sqrt{a + 4\sqrt{3ax} + 12x} \\
& = 3(m^2 + 2mn + n^2) \\
& = 3(m+n)^2 \\
& = 3(\sqrt{a} + 2\sqrt{3x})^2
\end{align}
$
Which is definitely a different expression.
Is the book being correct or me? Thanks in advance for answering this secondary school level question. I am an adult trying to make up my broken mathematics since teenage.
Thanks,
JF
(Update)
I found my mistake rather stupid. Derivation should be:
$ \begin{align} \sqrt{9a + 36\sqrt{3ax} + 108x} & = 3\sqrt{a + 4\sqrt{3ax} + 12x} \\ & = 3\sqrt{m^2 + 2mn + n^2} \\ & = 3\sqrt{(m+n)^2} \\ & = 3(m+n) \\ & = 3(\sqrt{a} + 2\sqrt{3x}) \\ & = 3\sqrt{a} + 6\sqrt{3x} \end{align} $
Which is same as the book's answer. Maybe Friday night isn't good time to re-picking up math. But I will keep on.
do $(3\sqrt{a}+6\sqrt{3x})^2$ and you get $9a+36\sqrt{3ax}+108x$
$\therefore \sqrt{9a+36\sqrt{3ax}+108x} = \sqrt{(3\sqrt{a}+6\sqrt{3x})^2} = |3\sqrt{a}+6\sqrt{3x}|$
Since $3\sqrt{a}+6\sqrt{3x}$ is always greater than zero, $|3\sqrt{a}+6\sqrt{3x}|=3\sqrt{a}+6\sqrt{3x}$
Hence it is true.