What is the answer to |x^2-2|+|x-1|=|x^2-x-1| ?

The answer to |x^2-2|+|x-1|=|x^2-x-1| is x<=-sqrt(2)\lor 1<= x<= sqrt(2)

Solve for x: |x^2-2|+|x-1|=|x^2-x-1|

The solution is x<=-sqrt(2)\lor 1<= x<= sqrt(2)

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The solution is x<=-sqrt(2)\lor 1<= x<= sqrt(2)

x^{2} - 2 + ∣ x - 1 ∣ = x^{2} - x - 1

x \leq - 2 or 1 \leq x \leq 2

Solution steps

x^{2} - 2 + ∣ x - 1 ∣ = x^{2} - x - 1

Find positive and negative intervals

x \leq - 2, - 2 < x \leq \frac{- 5 + 1}{2}, \frac{- 5 + 1}{2} < x < 1, 1 \leq x < 2, 2 \leq x < \frac{5 + 1}{2}, x \geq \frac{5 + 1}{2}

Solve the inequality for each interval

x \leq - 2 or No Solution or 1 \leq x < 2 or x = 2

Merge Overlapping Intervals

x \leq - 2 or 1 \leq x \leq 2

∣ x - 3 ∣ - ∣ 2 x - 5 ∣ = 0

\frac{1}{2} x - \frac{2}{3} - 1 = \frac{1}{6}

∣ x - 3 ∣ + ∣ 3 - x ∣ = 1

∣ x + 1 ∣ = 2 x - 3

∣ - x + 5 ∣ = 8

What is the answer to |x^2-2|+|x-1|=|x^2-x-1| ?
The answer to |x^2-2|+|x-1|=|x^2-x-1| is x<=-sqrt(2)\lor 1<= x<= sqrt(2)
Solve for x: |x^2-2|+|x-1|=|x^2-x-1|
The solution is x<=-sqrt(2)\lor 1<= x<= sqrt(2)