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And viceversa It should be tha same, because sqrt(x)+sqrt(x) = sqrt(2x) If not, why?

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Related. – Git Gud Apr 27 '14 at 15:21
Whether it’s a rule or a “rule”, you should always try it out with numbers to see whether it makes sense. Mathematics really is an observational science. – Lubin Apr 27 '14 at 15:50

$\sqrt{x} + \sqrt{x} = 2 \sqrt{x} = \sqrt{4x}$

$\sqrt{a} + \sqrt{b} \neq \sqrt{a+b}$ unless one of $a$ or $b$ is equal to zero.

"Regular" arithmetic doesn't work like that when dealing with square roots.

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$$\sqrt{3-x}-\sqrt x=\sqrt{3-2x}\iff 3-2\sqrt{3x-x^2}=3-2x\iff 3x-x^2=x^2\iff$$

$$\iff0=2x^2-3x=2x\left(x-\frac32\right)\iff x=0\;\;\;\text{or}\;\;\;x=\frac35$$

So no: it is not an equality.

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No, this is NOT true, for example $\sqrt2 \neq \sqrt1 + \sqrt1 = 2$

In general it is NOT true that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$

Edit: If it were true that $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ then by squaring up we would get

$$ a + b = a + 2\sqrt{ab} + b$$ so then $$2\sqrt{ab} = 0 \implies a = 0 \text{ or } b = 0 $$


$$ \sqrt{a+b} = \sqrt{a} + \sqrt{b} \iff a = 0 \text{ or } b = 0 $$

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