Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My understanding would lead me to believe that:

$$\frac{1}{2} \sqrt{4 + 4e^4} = \frac{1}{2}(2 + 2e^4) = 1 + e^4$$

But it actually equals: $\sqrt{1 + e^4}$

Can you explain why?

share|cite|improve this question
up vote 3 down vote accepted

You have $$ \frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{4}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4} $$ It looks like you where thinking that $$ \sqrt{a + b} = \sqrt{a} + \sqrt{b}. $$ But that is not true (try to check this with $a=b=2$). And even if you did that it looks like you forgot that $\sqrt{4e^4}= 2 e^2$ (as pointed out by TMM in the comment below.)

share|cite|improve this answer
+1. Lol! Exact same steps as mine! Interesting :) Deleted mine. – user17762 Mar 6 '13 at 19:18
@Marvis: No need to delete yours :) – Thomas Mar 6 '13 at 19:19
Even if he used $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ he should have had $1 + e^2$ instead of $1 + e^4$. – TMM Mar 6 '13 at 19:19
@TMM: That I think several rules were broken :) – Thomas Mar 6 '13 at 19:21

Also you can use squaring both sides: $$ \sqrt{4+4e^4}=2\sqrt{1+e^4}\\ 4+4e^4=4(1+e^4) $$ since $\sqrt{x^2}= \pm x$ and the expressions you start with is one of these two solutions (positive)

share|cite|improve this answer

Your mistake is to say that $\sqrt{4(a+b)}=2(a+b)$, which is not true. For instance $\sqrt{4(3^2+4^2)}=10\not=2(3^2+4^2)$

The correct equalities you need is $$\sqrt{(c^2a+c^2b)^2}=\sqrt{c^2(a+b)}=\sqrt{c^2}\sqrt{a+b}=c\sqrt{a+b}$$

Now $\frac{1}{2}\sqrt{4 + 4e^4} = \frac{1}{2}\sqrt{4(1+e^4)} = \frac{1}{2}\sqrt{2^2(1+e^4)} = \frac{1}{2}\sqrt{2^2}\sqrt{1 + e^4} = \frac{1}{2}2\sqrt{1+e^4} = \sqrt{1+e^4}$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.