# Thread: negative root?

1. ## negative root?

Why is it when you square root a number you get a +/-? Why is it when you solve $\displaystyle -4^2$ it is -16 and not 16? How can you tell it is +/- for the second question?

2. Hi there

$\displaystyle -4^2=-(4)(4)=-16$

but

$\displaystyle (-4)^2=(-4)(-4)=16$

You get a $\displaystyle \pm$ when you take the square root of a number because both the positive and negative square roots could be the factors: e.g. $\displaystyle 4\times4=16$ and $\displaystyle (-4)\times(-4)=16$

3. So, if it is not in ()'s I assume it is -(4)(4)? Why is it then in some cases it is always $\displaystyle -4^2$ is always 16?

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

$\displaystyle (4)^2=16,\ -(4^2)=-16$

$\displaystyle (-4)^2=16,\ -(-4)^2=-16$

$\displaystyle 4(4)=4+4+4+4=16$

$\displaystyle -4(4)=-4-4-4-4=-16$

$\displaystyle Hence,\ negative\ by\ positive\ means\ successive\ subtraction.$

$\displaystyle (-4)(-4)=-[4(-4)]=\ opposite\ of\ subtract\ 4\ four\ times,$

$\displaystyle which\ is\ add\ 4\ four\ times.\ Add\ and\ subtract\ are\ opposites.$

$\displaystyle This\ is\ why\ negative\ by\ negative\ is\ really\ positive\ by\ positive.$

$\displaystyle To\ write\ -4^2\ is\ ambiguous.$

$\displaystyle We\ must\ distinguish\ between\ -[(4)^2]\ and\ [(-4)^2]$

5. Originally Posted by Barthayn
So, if it is not in ()'s I assume it is -(4)(4)? Why is it then in some cases it is always $\displaystyle -4^2$ is always 16?
In my studies $\displaystyle -4^2$ has always been taken to equal $\displaystyle -16$. Maybe sometimes people get lazy and leave out the parentheses, or don't see a distinction between the two...