You learn in algebra that $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$

However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.


  • $\begingroup$ Note that the first two identities only hold when $a, b \geq 0$. As for your question, it depends on what one means by "rule", but the short answer is that there is no comparably simple expression. Since $\sqrt{\cdot}$ is concave, one can write down inequalities satisfies by $\sqrt{a + b}$. $\endgroup$ Jun 19, 2015 at 2:00
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    $\begingroup$ I suppose it depends on what you mean by a "rule". If you mean "some formula that lets you express $\sqrt{a+b}$ as some sort of combination involving $\sqrt a$ and $\sqrt b$", then no, there is no rule. $\endgroup$
    – mweiss
    Jun 19, 2015 at 2:01
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    $\begingroup$ @mweiss Of course, one can write $\sqrt{a + b} = \sqrt{(\sqrt{a})^2 + (\sqrt{b})^2}$, but this is hardly illuminating and really just emphasizes your point that the answer depends on what 'rule' means. $\endgroup$ Jun 19, 2015 at 2:04
  • $\begingroup$ @Travis : $\;\;\;$ More generally, they hold when $\: \operatorname{Im}\hspace{.02 in}(a) = 0 \leq \operatorname{Re}(a) \:$ or $\: \operatorname{Im}\hspace{.02 in}(b) = 0 \leq \operatorname{Re}(b) \;$. $\hspace{.88 in}$ $\endgroup$
    – user57159
    Jun 19, 2015 at 2:19
  • $\begingroup$ I suppose you could use imaginary numbers if you wanted... $\endgroup$
    – Kbot
    Jun 19, 2015 at 2:44

3 Answers 3


There is no finite rule but there is the binomial series: $$ (1 + x)^\alpha = \sum_{k=0}^{\infty} \; {\alpha \choose k} \; x^k $$ where $$ {\alpha \choose k} = \frac{\alpha (\alpha-1) (\alpha-2) \cdots (\alpha-k+1)}{k!} $$ This series converges for $|x|<1$. Taking $\alpha=1/2$, we get $$ (1+x)^{1/2} = 1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+\cdots $$

We can apply this to $\sqrt{a+b}$ as follows. Assume wlog that $a>b$. Then $$ \sqrt{a+b}=(a+b)^{1/2} = \sqrt{a}(1+x)^{1/2} $$ with $x=b/a<1$. We then have $$ \sqrt{a+b} = \sqrt{a}\left(1+\frac{b}{2a}-\frac{b^2}{8a^2}+\frac{b^3}{16a^3}-\frac{5 b^4}{128a^4}+\frac{7 b^5}{256a^5}+\cdots\right) $$

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    $\begingroup$ And if $b/a<1$, you can do it for with $x=a/b$. $\endgroup$
    – fonini
    Jun 19, 2015 at 2:32

Provided that $a>0$ and $a+b>0$, we have $$\sqrt{a+b}=\sqrt{a}\cdot\sqrt{1+\frac{b}{a}}$$

  • $\begingroup$ It's just like rewriting $a$ as $1\cdot a$ $\endgroup$
    – Vim
    Jun 19, 2015 at 10:07
  • $\begingroup$ This answer kind of reminds me of your answer here math.stackexchange.com/questions/454333/… I am a fan of your work! $\endgroup$
    – graydad
    Jun 24, 2015 at 1:35

The general answer is no. However, you may encounter some special occasions like: $$\sqrt{3+2\sqrt 2}=1+\sqrt 2$$ in which $\sqrt{3+2\sqrt 2}$ happens to be $(1+\sqrt 2)^2$.


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