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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.

jimfan
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  • The first line $9a + 36\sqrt{3ax} + 108x = 3\sqrt{a + 4\sqrt{3ax} + 12x}$ looks already wrong. – And you can simply check your result by calculating $(3\sqrt{a} + 6\sqrt{3x})^2$. – Martin R Feb 26 '21 at 15:28
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    To verify if the proposed answer is right or wrong, just square $3\sqrt{a} + 6\sqrt{3x}$. The book's answer squared does come out to $9a + 36\sqrt{3ax} + 108x$. So it is the square root you want. Your error lies in the first line of your calculations. An expression is not equal to its square root. – MasB Feb 26 '21 at 15:28
  • Thanks a lot. I updated the post. The book is indeed correct. – jimfan Feb 26 '21 at 16:01

1 Answers1

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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.

Sid
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