(Sorry if my MathJax is strange, I just skimmed through the tutorial and tried to make it work)
I want to find what is basically a sum formula for square roots, similar how it exists for $\log(a) + \log(b)\text{, }\sin(a) + \sin(b)$ and $\cos(a) + \cos(b)$ but I am looking for $\sqrt{a} + \sqrt{b}\space$. I am in pre-calc so I was wondering why such a formula was not already taught to me.
Ex. $\sqrt{2} + \sqrt{5}\space$ = some combination of radicals.
I tried it twice. The first try: $$\sqrt{a} + \sqrt{b}\space$$ $$= \sqrt{a} + \sqrt{a*\frac ba}\space$$ $$= \sqrt{a} + \sqrt{a}*\sqrt{\frac ba}\space$$ $$= \sqrt{a}(1 + \sqrt{\frac ba})\space$$ But I didn't really get anywhere with that. So I tried it a second time with a different method.
Let $x = \sqrt{a} + \sqrt{b}\space$. Then I solved a system of equations w/ difference of squares and some FOIL. $$x(\sqrt{a} - \sqrt{b}) = a-b$$ $$x(\sqrt{a} + \sqrt{b}) = a + 2\sqrt{ab} + b$$ Adding the equations, $$2x\sqrt{a} =2a + 2\sqrt{ab}$$ $$x\sqrt{a} =a + \sqrt{ab}$$ But I said only positive real numbers in the title, so squaring both sides should be fine. $$ax^2 = a^2 + 2a\sqrt{ab} + ab$$ $$x^2 = a + 2\sqrt{ab} + b$$ $$x = \sqrt{a + 2\sqrt{ab} + b}$$ $$\sqrt{a} + \sqrt{b}\space = \sqrt{a + 2\sqrt{ab} + b}$$ And it works, it eliminates the addition of square roots by forming a product. But I'm not sure if this was the most efficient way to solve it or if it works for all positive reals. Using the example, it works. $\sqrt{2} + \sqrt{5}\space$ = $\sqrt{7 + 2\sqrt{10}}$ and that is true, both sides are about 3.65. Is this a correct formula or am I missing something?