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(a+b)^1/2 and Square root (-4)^2=? I'm new to learning algebra. I know what (a+b)^2 is. But then I thought what happens with ^1/2 or ^1/4. Can someone explain me?

Also I have 2 questions in my book. Square root of 4^2= I calculated it by multiplying by ^1/2= 2/2=1, so I get 4^1 is 4. Then I got a question: Square root of (-4)^2. Doing the same steps here I ended up with -4. Is that correct?

I'm sorry if it is not wel explained. If something is unclear I'll try explaining diffrent.

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  • $\begingroup$ proofwiki.org/wiki/Exponent_Combination_Laws and see also : en.wikipedia.org/wiki/Principal_value $\endgroup$ Commented Aug 9, 2014 at 16:47
  • $\begingroup$ Since $(-4)^2 = 16$, then the ordinary, real-number square root $((-4)^2)^{1/2} = 16^{1/2} = \sqrt{16} = 4$. $\endgroup$
    – murray
    Commented Aug 9, 2014 at 16:48
  • $\begingroup$ I forgot the rule of BEDMAS. I should indeed have done brackets first. Thank you for that answer. $\endgroup$
    – user168852
    Commented Aug 9, 2014 at 16:51

3 Answers 3

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When you raise $x$ to the $\frac{1}{2}$ power, that means you are taking the square root of $x$. So $x^{\frac{1}{2}} = \sqrt{x}$. Similarly, $(a + b)^{\frac{1}{2}} = \sqrt{a + b}$.

Because of this, you can easily figure out what, for example, $4^{\frac{1}{2}}$ is. Since this is just $\sqrt{4}$, and you know that $\sqrt{4} = 2$, then we have $4^{\frac{1}{2}} = 2$.

Can you figure out what $9^{\frac{1}{2}}$ is? What about $16^{\frac{1}{2}}$?

Now, when you take the square root of anything, it is always a positive number. So if you have $\sqrt{ (-4)^{2}}$, first you square the $-4$ on the inside. This makes the problem become $\sqrt{16}$, and you know that $\sqrt{16} = 4$. This is why $\sqrt{ (-4)^{2}} = 4$. It's because you first square the inside, and the square root is always a positive number.

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Well you might have looked into

$\left(a+b\right)^2\:=\:a^2+b^2+2ab$

This is actually derived using BIOMIAL THEOREM. Similarly

$\left(a+b\right)^{\frac{1}{2}}\:\:or\:\left(a+b\right)^{\frac{1}{x}}$

for all real values of 'x' are derived using BINOMIAL THEOREM. If you are new to algebra BINOMIAL is a bit complex.

Actually we say $\left(a+b\right)^{\frac{1}{2}}$ as the square root of (a+b) written as $\sqrt{\left(a+b\right)}$.

And in the problem the first case is

$\left(4^2\right)^{\frac{1}{2}}=\left(16\right)^{\frac{1}{2}}$

Now $\left(16\right)^{\frac{1}{2}}$ is actually either +4 or -4 as both of them square yields 16.

That is

$4^2=\left(-4\right)^2=16$

So for the second part

$\left(\left(-4\right)^2\right)^{\frac{1}{2}}$

yields the same answer +4 or -4.

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  1. There is no rule for simplifying $\sqrt{a+b}$

  2. $(-4)^2 = -4 \cdot -4 = 16$

  3. $4^2$ cannot be calculated by applying an exponent of $\frac{1}{2}$ because it changes the question "what is the square of 4?"

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