# Curious about the plus/minus aspect of square roots

• Sep 27th 2010, 08:47 AM
Lord Darkin
Curious about the plus/minus aspect of square roots
Hello,

I understand that

$x^2=4$

Then $x= (+/-)2$

But what if I have

$x^2 = 4y^2$

Would that be x=2(plus/minus y) or would the -2 have to be included in the plus/minus part?
• Sep 27th 2010, 10:38 AM
emakarov
$x=\pm2$ is a contraction for "x=2 or x=-2". Even though $\pm y$ may seem a number, like y or -y, it cannot be viewed in isolation. Usually one has some proposition (i.e., an expression that is either true or false) $A(\pm y)$. This means " $A(y)$ or $A(-y)$". Therefore, " $x=2(\pm y)$" and " $x=\pm 2y$" are the equivalent and mean "x=2y or x = -2y".
• Sep 28th 2010, 05:02 AM
earboth
Quote:

Originally Posted by Lord Darkin
Hello,

I understand that

$x^2=4$

Then $x= (+/-)2$

But what if I have

$x^2 = 4y^2$

Would that be x=2(plus/minus y) or would the -2 have to be included in the plus/minus part?

To be pedantic $x = \pm 2$ is simply wrong because a number can't be positive and negative simultaneously. Compare emakarov's reply for that (She/he didn't use "and", I know).

1. Per definition the square-root is positive.

2. $x^2 = 4~\implies~|x| = 2$

3. Per definition the absolute value is

$|x|=\left\{\begin{array}{rcl}x&if&x\geq 0\\-x&if&x<0\end{array}\right.$

4. So

$x^2=4y^2~\implies~|x|=\left\{\begin{array}{rcl}2y& if&2y\geq 0\\-2y&if&2y<0\end{array}\right.$
• Sep 28th 2010, 05:10 AM
HallsofIvy
Quote:

Originally Posted by earboth
To be pedantic $x = \pm 2$ is simply wrong because a number can't be positive and negative simultaneously. Compare emakarov's reply for that (She/he didn't use "and", I know).

To be even more pedantic, $\pm$ does not mean "plus and minus", it means "plus or minus" so that $x= \pm 2$, meaning "x is either 2 or -2" is perfectly correct.

Quote:

1. Per definition the square-root is positive.

2. $x^2 = 4~\implies~|x| = 2$

3. Per definition the absolute value is

$|x|=\left\{\begin{array}{rcl}x&if&x\geq 0\\-x&if&x<0\end{array}\right.$

4. So

$x^2=4y^2~\implies~|x|=\left\{\begin{array}{rcl}2y& if&2y\geq 0\\-2y&if&2y<0\end{array}\right.$