Results 1 to 6 of 6

- September 6th 2011, 01:08 AM #1

- Joined
- Nov 2010
- Posts
- 19

## lxl = lyl

I wonna solve the equation lxl = lyl by using:

sqrt(x^2) = sqrt(y^2)

=>

x^2 = y^2

Here the easiest to do is to use the "zero-rule"

x^2 - y^2 = (x-y)(x+y) = 0

Where it's obvious that x=-y and y=-x etc.

However I don't understand why the same result isnt obtainable by not using this rule for solving the equation. What goes wrong when you do this:

x^2 = y^2

=>

x = (+-)sqrt(y^2)

Where you can't really get any further without comming back to the original statement. What is it that the zero-rules enables you in this case? Like when u factor out and use it instead of dividing by x, which if x=0 is not allowed algebraically.

- September 6th 2011, 01:37 AM #2

- September 6th 2011, 01:53 AM #3

- Joined
- Jul 2011
- Posts
- 212
- Thanks
- 2

- September 6th 2011, 02:13 AM #4

- Joined
- Nov 2010
- Posts
- 19

- September 6th 2011, 02:21 AM #5

- September 6th 2011, 05:10 AM #6

- Joined
- Apr 2005
- Posts
- 18,080
- Thanks
- 2393

## Re: lxl = lyl

Yes, you can. That says [tex]x= \pm|y|/tex] and, since |y| is either y or -y, you have x= y or x= -(y), x= (-y) or x= -(-y)= y giving the correct solutions. x= y or x= -y.

What is it that the zero-rules enables you in this case? Like when u factor out and use it instead of dividing by x, which if x=0 is not allowed algebraically.