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

The solution to |x^2-2x-2|>= 1 is x = 3

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

x^{2} - 2 x - 2 \geq 1

x \leq - 1 or - 2 + 1 \leq x \leq 2 + 1 or x \geq 3

Interval Notation

(- \infty, - 1] \cup [- 2 + 1, 2 + 1] \cup [3, \infty)

Decimal Notation

x \leq - 1 or - 0.41421 \dots \leq x \leq 2.41421 \dots or x \geq 3

Solution steps

x^{2} - 2 x - 2 \geq 1

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

x^{2} - 2 x - 2 \leq - 1 or x^{2} - 2 x - 2 \geq 1

x^{2} - 2 x - 2 \leq - 1 or x^{2} - 2 x - 2 \geq 1

- 2 + 1 \leq x \leq 2 + 1 or (x \leq - 1 or x \geq 3)

Merge Overlapping Intervals

x \leq - 1 or - 2 + 1 \leq x \leq 2 + 1 or x \geq 3

5 x^{2} + 3 x - 5 = 0

(x + 2) (x + 2) = 0

5 b^{2} + 21 b + 4

9^{\frac{5}{2}}

(f^{- 3} g^{5} h^{8})^{- 3}

What is |x^2-2x-2|>= 1 ?
The solution to |x^2-2x-2|>= 1 is x<=-1\lor-sqrt(2)+1<= x<= sqrt(2)+1\lor x>= 3