What is |x^2-4x+2|>= |x+1| ?

The solution to |x^2-4x+2|>= |x+1| is x = (sqrt(21)+5)/2

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|x^2-4x+2|>= |x+1|

x^{2} - 4 x + 2 \geq ∣ x + 1 ∣

x \leq \frac{- 21 + 5}{2} or x \geq \frac{21 + 5}{2}

Solution steps

x^{2} - 4 x + 2 \geq ∣ x + 1 ∣

Find positive and negative intervals

x < - 1, - 1 \leq x \leq - 2 + 2, - 2 + 2 < x < 2 + 2, x \geq 2 + 2

Solve the inequality for each interval

x < - 1 or - 1 \leq x \leq \frac{- 21 + 5}{2} or No Solution or x \geq \frac{21 + 5}{2}

Merge Overlapping Intervals

x \leq \frac{- 21 + 5}{2} or x \geq \frac{21 + 5}{2}

∣ u - 5 ∣ \leq 5

∣ 3 - 2 x ∣ < 4

∣ x + 3 ∣ > 7

1 lo g_{3} (∣ 3 - 4 x ∣) > 2

∣ x + 2 ∣ + ∣ 3 x + 5 ∣ \geq 2

What is |x^2-4x+2|>= |x+1| ?
The solution to |x^2-4x+2|>= |x+1| is x<= (-sqrt(21)+5)/2 \lor x>= (sqrt(21)+5)/2