What is |x^2-x|<= x+1 ?

The solution to |x^2-x|<= x+1 is -sqrt(2)+1<= x<= sqrt(2)+1

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|x^2-x|<= x+1

x^{2} - x \leq x + 1

- 2 + 1 \leq x \leq 2 + 1

Interval Notation

[- 2 + 1, 2 + 1]

Decimal Notation

- 0.41421 \dots \leq x \leq 2.41421 \dots

Solution steps

x^{2} - x \leq x + 1

Find positive and negative intervals

x \leq 0, 0 < x < 1, x \geq 1

Solve the inequality for each interval

- 2 + 1 \leq x \leq 0 or 0 < x < 1 or 1 \leq x \leq 2 + 1

Merge Overlapping Intervals

- 2 + 1 \leq x \leq 2 + 1

∣ 2 x - 3 ∣ > ∣ x + 4 ∣

∣ - 0.5 x ∣ < 1

[\frac{( x ^{2} - 5 x + 6 )}{( ∣ x ∣ + 7 )}] < 0

∣ 4 - 3 x ∣ - 2 > 0

∣ 3 x - 1 ∣ > 2 x

What is |x^2-x|<= x+1 ?
The solution to |x^2-x|<= x+1 is -sqrt(2)+1<= x<= sqrt(2)+1