# Square Root

• Jul 18th 2008, 05:45 AM
magentarita
Square Root
Why is the answer to a square root problem always positive and negative?

SAMPLE:

sqrt{16} = -4 and +4

• Jul 18th 2008, 05:58 AM
galactus
Because $\displaystyle (-4)^{2}=16$ and $\displaystyle 4^{2}=16$
• Jul 18th 2008, 05:59 AM
Simplicity
Quote:

Originally Posted by magentarita
Why is the answer to a square root problem always positive and negative?

SAMPLE:

sqrt{16} = -4 and +4

$\displaystyle (-4)^2 = 16$
$\displaystyle 4^2 = 16$
$\displaystyle \implies \sqrt{16} = \pm 4$

Visually it may appear clearer, consider the quadratic graph of $\displaystyle y = x^2$ (Shown Below), for every $\displaystyle \pm x$ value, it is mapped onto the same $\displaystyle y$ value.
• Jul 18th 2008, 06:59 AM
magentarita
I thank both replies. I want to thank Air for the picture, which makes the answer clearer.

By the way, does the same applies to variables?

For exmaple:

sqrt{x^2} = -x and +x??? True or false?
• Jul 18th 2008, 09:37 AM
Simplicity
Quote:

Originally Posted by magentarita
I thank both replies. I want to thank Air for the picture, which makes the answer clearer.

By the way, does the same applies to variables?

For exmaple:

sqrt{x^2} = -x and +x??? True or false?

True. (Clapping)

Variables are just numbers in disguise. (A range of constants)

$\displaystyle \sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x \le 0 \end{cases}$
• Jul 18th 2008, 09:37 AM
Plato
I agree that there are two square roots of any positive real number.

But, I disagree completely that $\displaystyle \sqrt {16} = \pm 4$. That is simply an abuse of notation!

This is a standard discussion in any elementary mathematics course.
The two square roots of 16 are: $\displaystyle \sqrt {16} = 4\,\& \, - \sqrt {16} = - 4$.

Therefore, $\displaystyle \sqrt {x^2 } = \left| x \right|$.
• Jul 18th 2008, 10:31 AM
Moo
Quote:

Originally Posted by Plato
I agree that there are two square roots of any positive real number.

But, I disagree completely that $\displaystyle \sqrt {16} = \pm 4$. That is simply an abuse of notation!

This is a standard discussion in any elementary mathematics course.
The two square roots of 16 are: $\displaystyle \sqrt {16} = 4\,\& \, - \sqrt {16} = - 4$.

Therefore, $\displaystyle \sqrt {x^2 } = \left| x \right|$.

Completely agree with that !

If you see the graph of the function y=sqrt(x), you will see that y can't be negative.

Actually, working with the graph y=x² is a mistake because sqrt(x) is not the inverse function of x² !!!!

If one has x²=16, then for sure x=+ or - sqrt(16) because x²=16 --> x²-16=0 --> x²-(sqrt(16))²=0 --> (x-sqrt(16))(x+sqrt(16))=0 and the rest follows.

Quote:

Originally Posted by Air
True. (Clapping)

Variables are just numbers in disguise. (A range of constants)

$\displaystyle \sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x \le 0 \end{cases}$

Quote:

Originally Posted by Air
$\displaystyle (-4)^2 = 16$
$\displaystyle 4^2 = 16$
$\displaystyle \implies \sqrt{16} = \pm 4$
Visually it may appear clearer, consider the quadratic graph of $\displaystyle y = x^2$ (Shown Below), for every $\displaystyle \pm x$ value, it is mapped onto the same $\displaystyle y$ value.