(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.


Well you might have looked into


This is actually derived using BIOMIAL THEOREM. Similarly


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


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

That is


So for the second part


yields the same answer +4 or -4.


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.

  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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