# Absolute value and square root

## Main Question or Discussion Point

what is /squared root sign4x+1 /
what does this equal because i'm confused when it has the square root sign on it all and it's absolute value

cristo
Staff Emeritus
Do you mean $|\sqrt{4x+1}|$? Here, the absolute value sign just means take the positive square root.

yup that's what i mean

so when i a take the absolute value of it, it looks the same or the square root is gone

cristo
Staff Emeritus
The square root is still there, since that is the operation being applied to (4x+1). When we take the square root of a number, we get a positive number, and a negative number (consider the simple example: $\sqrt{4}=\pm 2$, since (-2)2=4=22). By putting the absolute value around the square root is the same as saying that we are taking the positive square root (so, in our example$|\sqrt{4}|=+\sqrt{4}=2$).

Your expression above can be written $+\sqrt{4x+1}$.

what happens after i take the absolut value of square root sign4x+1

so that's what it'll look like

Your expression above can be written $+\sqrt{4x+1}$.
Actually, his expression can be written $\sqrt{4x+1}$. Square roots are defined to be a FUNCTION, which means they CAN'T give you more than one result for any number in their domain (i.e. we can't have $\sqrt{4}=\pm 2$). By convention, $\sqrt{x} \ge 0$ for all $x \in [0,\infty)$.

cristo
Staff Emeritus
Actually, his expression can be written $\sqrt{4x+1}$. Square roots are defined to be a FUNCTION, which means they CAN'T give you more than one result for any number in their domain (i.e. we can't have $\sqrt{4}=\pm 2$). By convention, $\sqrt{x} \ge 0$ for all $x \in [0,\infty)$.
Lets be clear that in general |f(x)| can't be written as simply +f(x)! Here that was true because $\sqrt{x}$ is by definition non-negative. If f(x)= x and x= -4 then |f(x)|= |-4|= 4 while +f(x)= +(-4)= -4.