Simplify $\sqrt{7+4\sqrt{3}}+\sqrt{28-10\sqrt{3}}$

Question

I have this expression:

$\sqrt{7+4\sqrt{3}}+\sqrt{28-10\sqrt{3}}$

I know how to simplify it completing the square inside the "big" square roots but my questions is if there is another way (easier or not) to get the same result.

Answer 1

I wouldn't say that it's "obvious" that $7 + 4\sqrt{3} = (2 + \sqrt{3})^2$. Certainly it's obvious once someone tells you (you can expand out the right-hand side to get the left-hand side), but how would you derive it?

Say you wanted to derive that - that is, to find integers $a$ and $b$ such that $$7 + 4\sqrt{3} = (a + b\sqrt{3})^2$$ if those integers exist. Expanding out the right-hand side we get $$(a^2 + 3b^2) + 2ab \sqrt{3}$$ and so we need to find $a, b$ such that $a^2 + 3b^2 = 7, 2ab = 4$. So $ab = 2$ and either $a = 2, b = 1$ or vice versa; it turns out that $a = 2, b = 1$ works to make $a^2 + 3b^2 = 7$.

If on the other hand we wanted to find $\sqrt{7 + 6 \sqrt{3}}$, we'd look for $a, b$ such that $a^2 + 3b^2 = 7, 2ab = 6$. Now either $a = 3, b = 1$ or $a = 1, b = 3$ -- but neither of these satisfies $a^2 + 3b^2 = 7$, so $7 + 6\sqrt{3}$ doesn't have a square root of the form $a + b\sqrt{3}$ for integers $a, b$. (Formally, it's not an element of $\mathbb{Z}[\sqrt{3}]$; if you don't know what this means, ignore it.)

Answer 2

HINT:

$$7+4\sqrt{3}=(2+\sqrt{3})^2$$ $$28-10\sqrt{3}=(5-\sqrt{3})^2$$

Answer 3

You don't have to complete any square: it is obvious that

$$7+4\sqrt 3=(2+\sqrt3)^2, \quad 28-10\sqrt3=(5-\sqrt3)^2.$$