$|x^2 -x| - |1-x| = 0$

Question

$$|x^2 -x| - |1-x| = 0$$

I'm trying to find the values that $x$ can take by thinking case by case like $x<1 $ or $x>1$. How do you apply it for this question?

Regards

Answer 1

$|x^2 - x| - |1-x| = 0$

$|x^2 -x| = |1-x|$

$|x||x-1| = |x-1|$

so either i) $|x - 1|= 0$ or ii) $|x| = \frac {|x-1|}{|x-1|} = 1$ and $|x-1|\ne 0$

If i) then $|x-1| = 0$ so $x-1 = 0$ and $x = 1$.

If ii) then $|x| =1$ and $x = \pm 1$ (while $|x-1| \ne 0$ so $x \ne 1$ so $x = -1$)

So either $x = 1$ or $x = -1$.

I was going to suggest you can always do absolute values in cases:

1) $x^2 - x \ge 0$ and $1 - x \ge 0$ so $(x^2 - x)-(1-x) = x^2 - 1 = 0 $ so $x^2 = 1$ so $x = \pm 1$ which are both compatible with $1-x \ge0$ and $x^2 - x \ge 0$. so $x =\pm 1$

2) $x^2 - x \ge 0$ and $1- x < 0$ so $(x^2 - x) - (x-1) = x^2 -2x + 1 = (x-1)^2 = 0$ so $x-1 = 0$ so $x = 1$ but $1 < x$ so that's impossible.

3) $x^2 - x <0$ and $1 - x \ge 0$ so $(x - x^2) - (1-x) = -1 + 2x - x^2 =-(1-x)^2 = 0$ so $x=1$ but $x^2 - x < 0$ so that is impossible.

4) $x^2 - x < 0$ and $1-x < 0$ so $(x-x^2) - (x-1) = -x^2 + 1 = 0$ so $x = \pm 1$ but $1 < x$ and $x^2 < x$ so those are impsossible.

... but the absolute diresctness of $|x^2 - x| = |x||x-1| = |x-1|$ was just to straightforward and simple to ignore.

Answer 2

Hint:   write it as $|x| \cdot |x-1| - |1-x| = 0 \iff (|x|-1) \cdot |x-1| = 0\,$

Answer 3

The crucial points are $0,1$. If $x\geq 1$, then $|x(x-1)|=x(x-1)$, $|x-1|=x-1$. If $0\leq x<1$, then $|x(x-1)|=-x(x-1)$, $|x-1|=-(x-1)$. If $x<0$, then $|x(x-1)|=x(x-1)$, $|x-1|=-(x-1)$.

Answer 4

Simplify:

$|x^2-x| -|1-x|=0;$

$|x||x-1| -|1-x|=0;$

$|x-1|(|x| -1)=0;$

1)$ |x-1|=0$ , $x=1;$

2) $|x| -1 =0$ , $|x|=1$, $x= \pm~1$.