Question

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

Answer 1

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Answer 2

To solve the absolute value equation |x^2 - 1| - |x^2 - 2x| = |2x - 1|, we need to consider different cases based on the sign of the expressions inside the absolute value bars.

Case 1: x^2 - 1 ≥ 0 and x^2 - 2x ≥ 0 and 2x - 1 ≥ 0 Solving each of these inequalities separately:

x^2 - 1 ≥ 0 x^2 ≥ 1 x ≥ 1 or x ≤ -1

x^2 - 2x ≥ 0 x(x - 2) ≥ 0 x ≤ 0 or x ≥ 2

2x - 1 ≥ 0 2x ≥ 1 x ≥ 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

x ≥ 1 (from the 2nd and 3rd inequalities) x ≤ 0 (from the 1st and 2nd inequalities)

However, these intervals do not overlap, so there are no solutions in this case.

Case 2: x^2 - 1 ≥ 0 and x^2 - 2x < 0 and 2x - 1 ≥ 0 Solving each of these inequalities separately:

x^2 - 1 ≥ 0 x^2 ≥ 1 x ≥ 1 or x ≤ -1

x^2 - 2x < 0 x(x - 2) < 0 0 < x < 2

2x - 1 ≥ 0 2x ≥ 1 x ≥ 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

1 ≤ x < 2 (from the 1st and 2nd inequalities) 1/2 ≤ x < 2 (from the 2nd and 3rd inequalities)

The common interval is 1 ≤ x < 2.

Case 3: x^2 - 1 ≥ 0 and x^2 - 2x ≥ 0 and 2x - 1 < 0 Solving each of these inequalities separately:

x^2 - 1 ≥ 0 x^2 ≥ 1 x ≥ 1 or x ≤ -1

x^2 - 2x ≥ 0 x(x - 2) ≥ 0 x ≤ 0 or x ≥ 2

2x - 1 < 0 2x < 1 x < 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

x < 1/2 (from the 1st and 3rd inequalities) x ≤ 0 (from the 1st and 2nd inequalities)

The common interval is x < 1/2.

Case 4: x^2 - 1 < 0 and x^2 - 2x < 0 and 2x - 1 ≥ 0 Solving each of these inequalities separately:

x^2 - 1 < 0 x^2 < 1 -1 < x < 1

x^2 - 2x < 0 x(x - 2) < 0 0 < x < 2

2x - 1 ≥ 0 2x ≥ 1 x ≥ 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

-1 < x < 1 (from the 1st and 2nd inequalities) 1/2 ≤ x (from the 2nd and 3rd inequalities)

The common interval is 1/2 ≤ x < 1.

Case 5: x^2 - 1 < 0 and x^2 - 2x ≥ 0 and 2x - 1 < 0 Solving each of these inequalities separately:

x^2 - 1 < 0 x^2 < 1 -1 < x < 1

x^2 - 2x ≥ 0 x(x - 2) ≥ 0 x ≤ 0 or x ≥ 2

2x - 1 < 0 2x < 1 x < 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

-1 < x < 1/2 (from the 1st and 3rd inequalities) x ≤ 0 (from the 1st and 2nd inequalities)

The common interval is -1 < x < 1/2.

Case 6: x^2 - 1 ≥ 0 and x^2 - 2x < 0 and 2x - 1 < 0 Solving each of these inequalities separately:

x^2 - 1 ≥ 0 x^2 ≥ 1 x ≥ 1 or x ≤ -1

x^2 - 2x < 0 x(x - 2) < 0 0 < x < 2

2x - 1 < 0 2x < 1 x < 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

x < 1/2 (from the 2nd and 3rd inequalities) x ≤ -1 (from the 1st and 2nd inequalities)

The common interval is x ≤ -1.

Case 7: x^2 - 1 < 0 and x^2 - 2x < 0 and 2x - 1 < 0 Solving each of these inequalities separately:

x^2 - 1 < 0 x^2 < 1 -1 < x < 1

x^2 - 2x < 0 x(x - 2) < 0 0 < x < 2

2x - 1 < 0 2x < 1 x < 1/2

Now, we need to find the intersection of these intervals to satisfy all three inequalities:

-1 < x < 1/2 (from the 1st and 3rd inequalities) x < 1/2 (from the 2nd and 3rd inequalities)

The common interval is -1 < x < 1/2.

After considering all cases, we can summarize the solutions to the original equation:

1 ≤ x < 2 1/2 ≤ x < 1 -1 < x < 1/2 x ≤ -1

So, the solution to the equation |x^2 - 1| - |x^2 - 2x| = |2x - 1| is x ≤ -1, -1 < x < 1/2, 1/2 ≤ x < 1, and 1 ≤ x < 2.