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I'd like to know how to solve the following square root:-

$\sqrt{4+4x}$

The result is: $2\sqrt{x+1}$

I did not understand where the $x+1$ come from.

rikusp2002
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Mark Cunnighan
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    As per the rules of indices, we can write $$\sqrt{ab} = \sqrt{a} * \sqrt{b}$$ So, $$\sqrt{4(x + 1)} = \sqrt{4} * \sqrt{x + 1}$$ and you can do the rest. – rikusp2002 May 28 '19 at 16:22
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    $\sqrt{4+4x} = \sqrt{4(1 + x)} = \sqrt{4}\sqrt{1+x} = 2 \sqrt{1+x}$ since $\sqrt{ab} = \sqrt{a}\sqrt{b}$ whenever at least $a$ or $b$ is positive. – Jan May 28 '19 at 16:23
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    Is it so hard to see that $4+4x=4(1+x)$ ? –  May 28 '19 at 16:50

2 Answers2

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Just to note, we are not solving anything since there is no equal sign; we are only simplifying: $$\sqrt{4+4x}=\sqrt{4(x+1)}=\sqrt{4}\cdot\sqrt{x+1}=2\sqrt{x+1}$$

I used the property that $$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$$

coreyman317
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    Note that your calculation is also valid if $x<-1$, since for the property $\sqrt{ab} = \sqrt{a} \sqrt{b}$ it suffices that $a$ or $b$ is positive. – Jan May 28 '19 at 16:27
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For positive $a$ and $b$, $\sqrt{ab}=\sqrt a\sqrt b$. $$\sqrt{4+4x}=\sqrt{4(x+1)}=\sqrt4\sqrt{x+1}=2\sqrt{x+1}$$

Parcly Taxel
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