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?

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- Jan 1st 2010, 04:59 AMBarthaynnegative 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?

- Jan 1st 2010, 05:15 AMStroodle
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 $\displaystyle (-4)\times(-4)=16$ - Jan 1st 2010, 05:18 AMBarthayn
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?

- Jan 1st 2010, 05:19 AMArchie Meade
$\displaystyle -(4)^2=-(-4)^2=-16$

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

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

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

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

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

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

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

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

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

$\displaystyle We\ must\ distinguish\ between\ -[(4)^2]\ and\ [(-4)^2]$ - Jan 1st 2010, 05:23 AMStroodle