Square root of surds: $\sqrt{12+2\sqrt{6}}$?

Question

I got this question

Find the square root of $12+2\sqrt{6}$ expressing your answer in the form $\sqrt{m}+\sqrt{n}$.

I have no idea what this means and how to go about it.

It was a serious question. In 7 years of undergrad and graduate work, I've never seen that word. I googled it, but I am fairly confident that is an unusual word. Someone correct me if I'm wrong. — Euler....IS_ALIVE, Feb 03 '13 at 23:06
@Euler....IS_ALIVE Well, since you know what I imply by surds, can you please tell me your own term for it? — Mob, Feb 03 '13 at 23:16
@Mob Irrational numbers, probably. (Although that's a broader term.) — , Mar 05 '14 at 14:51
See also: Denesting radicals at Wikipedia, Strategies to denest nested radicals (and the posts linked there), Denesting a square root: $\sqrt{7 + \sqrt{14}}$ (and the posts linked there) — Martin Sleziak, Oct 16 '16 at 09:15

score 9 · Accepted Answer · answered Feb 03 '13 at 23:06

9

Let $x = \sqrt{12 + 2\sqrt{6}} = \sqrt{n} + \sqrt{m}$. Then $x^2 = 12 + 2\sqrt 6 = n + m + 2 \sqrt{nm}$.

Find $n$ and $m$ such that $n + m = 12$ and $nm = 6$.

answered Feb 03 '13 at 23:06

Damien L

Can you expatiate on this? – Mob Feb 03 '13 at 23:13
2

Surdly you can solve the system of equations Damien has given. – Euler....IS_ALIVE Feb 03 '13 at 23:18

score 1 · Answer 2 · answered Mar 02 '15 at 12:24

1

The answers for $n$ and $m$ are

$n = 6+\sqrt{30}$

$m = 6-\sqrt{30}$.

Notice that they are interchangable.

answered Mar 02 '15 at 12:24

instcff

2 Answers2