- #1
- 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?