Absolute value and square root

[afcwestwarrior]

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]

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

 

[afcwestwarrior]

yup that's what i mean

 

[afcwestwarrior]

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

 

[cristo]

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 $+\sqrt{4x+1}$.

 

[afcwestwarrior]

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

 

[afcwestwarrior]

so that's what it'll look like

 

[Moo Of Doom]

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]

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 $x \in [0,\infty)$.

Good point; thanks for spotting that, Moo!

 

[HallsofIvy]

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.