- #1

greg_rack

Gold Member

- 363

- 79

- Homework Statement
- $$\sqrt{x^{2}-2x}-x+1>0$$

- Relevant Equations
- none

I got this function in a function analysis and got confused on how to solve its positivity;

I rewrote it as:

$$\sqrt{x^{2}-2x}>x-1 \rightarrow x^2-2x>x^2-2x+1$$

And therefore concluded it must've been impossible... but I'm certainly missing something stupid, since plotting the graphs of the two functions(##\sqrt{x^{2}-2x}## and ##x-1##) I see that the first is greater only for ##x<0##.

Maybe the flaw comes when I'm squaring both factors in the inequality... should I put the first factor in an absolute value, since it must be positive, as a square root?

But then, how do I solve it with the abs.?

I can deduce that the left term must be greater until the second term is negative... and for positive values it's the reversed situation, and this thinking works, but is there a more "rigid" procedure?

I rewrote it as:

$$\sqrt{x^{2}-2x}>x-1 \rightarrow x^2-2x>x^2-2x+1$$

And therefore concluded it must've been impossible... but I'm certainly missing something stupid, since plotting the graphs of the two functions(##\sqrt{x^{2}-2x}## and ##x-1##) I see that the first is greater only for ##x<0##.

Maybe the flaw comes when I'm squaring both factors in the inequality... should I put the first factor in an absolute value, since it must be positive, as a square root?

But then, how do I solve it with the abs.?

I can deduce that the left term must be greater until the second term is negative... and for positive values it's the reversed situation, and this thinking works, but is there a more "rigid" procedure?