# How does $\frac{1}{2} \sqrt{4 + 4e^4} = \sqrt{1 + e^4}$

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?

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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.)

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+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)

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

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