negative root?

**Barthayn** · January 1st 2010, 04:59 AM

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

**Stroodle** · January 1st 2010, 05:15 AM

Hi there

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

but

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

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

**Barthayn** · January 1st 2010, 05:18 AM

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

**Archie Meade** · January 1st 2010, 05:19 AM

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

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

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

$4(4)=4+4+4+4=16$

$-4(4)=-4-4-4-4=-16$

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

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

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

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

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

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

**Stroodle** · January 1st 2010, 05:23 AM

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 $-4^2$ is always 16?

In my studies $-4^2$ has always been taken to equal $-16$ . Maybe sometimes people get lazy and leave out the parentheses, or don't see a distinction between the two...

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