# Solutions to x

1. Nov 9, 2012

### christian0710

This may be a dumb question but if we have the equation 2x=x2 and we use algebra we get
2=x2/x ---> 2=x

How come the solution is x= 2 (obvious) AND x=0 (not so obvious for me)

2. Nov 9, 2012

### haruspex

Clearly x=0 is a valid solution of the original equation. But the important lesson is that whenever you divide by an expression of unknown value (x in this case) you should bear in mind that division by zero is not a defined operation. The correct procedure is always to write "if (expression) is nonzero then ...".

3. Nov 9, 2012

### christian0710

I see so you obtain that if x=0 you get 2*0=^2 and that's why x=0 is an equation?
But if we started with the equation x=2, then i assume you can't say x=2 and x=0, is that correctly understood? I needed both solutions because i was finding the upper and lower limits for integration, but just confues about the fact that x=2 also has x=0 as solution.

4. Nov 9, 2012

### mtayab1994

Well doesnt 2x=x^2 ⇔0=x^2-2x ⇔ 0=x(x-2) ⇔x=0 or x=2 Is that clear now? It is pretty simple.

5. Nov 9, 2012

### Staff: Mentor

If x = 0, you get 2*0 = 02, so x = 0 is a solution to the original equation.
The only possible replacement for x in the equation x = 2 is 2. That's the only value that makes the equation x = 2 a true statement.
The equation x = 2 does NOT have x = 0 as a solution. The equation 2x = x2 DOES have x = 0 (and x = 2) as a solution.

Since the equations x = 2 and 2x = x2 have different solution sets, they are not equivalent.

6. Nov 9, 2012

### christian0710

Thank you so much, now it's clear! Very clear :D

7. Nov 9, 2012

### mtayab1994

You're very welcome :D
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