# Thread: lxl = lyl

1. ## 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.

2. ## Re: lxl = lyl

Originally Posted by aaaa202
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.
Why do you even need to bother using this method? It's clear that the size of x is the same as the size of y, which means y = x or y = -x...

3. ## Re: lxl = lyl

the whole state of affairs can be described by the following chain: $\displaystyle |x|=|y|\Longleftrightarrow\sqrt {x^2} = \sqrt {y^2}\Longleftrightarrow$ $\displaystyle x^2=y^2\Longleftrightarrow (x+y)(x-y)=0\Longleftrightarrow x=-y\vee x=y$

4. ## Re: lxl = lyl

yeh okay. I just don't get why the method of solving x or y wont get you a result..

5. ## Re: lxl = lyl

Originally Posted by aaaa202
yeh okay. I just don't get why the method of solving x or y wont get you a result..
Well psolaki showed you that it can, I'm just saying that understanding the modulus function means you don't need to go through that rigmarole...

6. ## Re: lxl = lyl

Originally Posted by aaaa202
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.
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.
I have no idea what you are asking here.

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# solve question lxl lyl 2

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