Need some help on my calc homework

[Xylitol]

This is only a small part of it, but even though it's basic, it's important

How do you simplify this:
SquareRoot[9x^2+x]/x

?
Thanks

 

[Leeroy]

Start by realising SQRT(x) = x^1/2, and 1/x = x^-1

 

[Saint Michael]

(9+1/x)^(1/2)

 

[Xylitol]

Originally posted by: Leeroy
Start by realising SQRT(x) = x^1/2, and 1/x = x^-1

I'm so rusty with this easy math that I'm so lost. How would I get to Saint Michael's answer using your info?

 

[Saint Michael]

The square root of a product of two factors is equal to the product of the square roots of those two factors. Therefore we restate (9(x^2)+x) as (x^2)(9 + 1/x). Therefore the square root of (9(x^2) + x) is equal to the square root of (x^2) times the square root of (9 + 1/x). The square root of (x^2) is x, we cancel numerator and denominator x and we are left with the square root of (9 + 1/x).

 

[Ricemarine]

http://calc101.com/webMathemat...erivatives.jsp#topdoit

It says to quotient rule... But I can see where Saint Michael is coming from...

 

[Saint Michael]

Originally posted by: Saint Michael
The square root of a product of two factors is equal to the product of the square roots of those two factors. Therefore we restate (9(x^2)+x) as (x^2)(9 + 1/x). Therefore the square root of (9(x^2) + x) is equal to the square root of (x^2) times the square root of (9 + 1/x). The square root of (x^2) is x, we cancel numerator and denominator x and we are left with the square root of (9 + 1/x).

Edit: Actually, I guess this is only that simple if x > 0. Otherwise you're looking at |x|(9 + 1/x)^(1/2)/x, which means that you're looking at (-1)(9 + 1/x)^(1/2) or (9 + 1/x)^(1/2) depending on whether x is negative or positive (x can't be 0). Hmm. Dunno if that's clean enough or not.

Real Edit: Dammit, meant for this to be an edit.