1
$\begingroup$

(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?

$\endgroup$
  • $\begingroup$ That is correct, but not of as much use as that of $\sin$'s or $\log$'s. $\endgroup$ – Trebor Apr 3 at 6:07
  • 1
    $\begingroup$ If $x$ is positive then $x=\sqrt {x^{2}}$ so it is obvious that $\sqrt a+\sqrt b =\sqrt {(\sqrt a+\sqrt b)^{2}}=\sqrt { a+2\sqrt {ab} +b}$. $\endgroup$ – Kavi Rama Murthy Apr 3 at 6:09
  • $\begingroup$ There is a formula for $\sin a+\sin b$? I have heard of $\sin(a+b)$, but not the other way around. Also, $\ln a+\ln b=\ln ab$ is completely natural, as translating between addition and multiplication is, is some sense, what logarithms are for. The fact that it's difficult to find something sensible for $\sqrt a+\sqrt b$ or $\sqrt{a+b}$ is just another testament to the fact that multiplication and addition often get in one another's way (see the difficulty we have with describing the primes, or Fermat's last theorem for other examples). $\endgroup$ – Arthur Apr 3 at 6:17
  • $\begingroup$ @Arthur Here's a trig reference sheet I have bookmarked. Check page 2, right column: "sum to product formulas". tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf $$\sin \alpha + \sin \beta = 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha - \beta}{2}$$ $\endgroup$ – Eevee Trainer Apr 3 at 6:19
  • 1
    $\begingroup$ @Arthur $\sin a+\sin b=2\sin\frac{a+b}2\cos\frac{a-b}2$. $\endgroup$ – Lord Shark the Unknown Apr 3 at 6:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.