What is |-x^2-9x-7|>= 1 ?

The solution to |-x^2-9x-7|>= 1 is x = (sqrt(57)-9)/2

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|-x^2-9x-7|>= 1

- x^{2} - 9 x - 7 \geq 1

x \leq \frac{- 57 - 9}{2} or - 8 \leq x \leq - 1 or x \geq \frac{57 - 9}{2}

Interval Notation

(- \infty, \frac{- 57 - 9}{2}] \cup [- 8, - 1] \cup [\frac{57 - 9}{2}, \infty)

Decimal Notation

x \leq - 8.27491 \dots or - 8 \leq x \leq - 1 or x \geq - 0.72508 \dots

Solution steps

- x^{2} - 9 x - 7 \geq 1

Apply absolute rule: If

∣ u ∣ \geq a, a > 0

then

u \leq - a

u \geq a

- x^{2} - 9 x - 7 \leq - 1 or - x^{2} - 9 x - 7 \geq 1

- x^{2} - 9 x - 7 \leq - 1 or - x^{2} - 9 x - 7 \geq 1

(x \leq \frac{- 57 - 9}{2} or x \geq \frac{57 - 9}{2}) or - 8 \leq x \leq - 1

Merge Overlapping Intervals

x \leq \frac{- 57 - 9}{2} or - 8 \leq x \leq - 1 or x \geq \frac{57 - 9}{2}

∣ x - 2 ∣ > 2 + x - ∣ 3 - x ∣

∣ x - 1 ∣ + ∣ x - 2 ∣ + ∣ x - 3 ∣ \leq 6

\frac{1}{∣ x - 2 ∣} \geq \frac{1}{4}

\frac{5}{( 3 - x )} > 1

4 ∣ x - 2 ∣ - 2 - x < 4 x + 1

What is |-x^2-9x-7|>= 1 ?
The solution to |-x^2-9x-7|>= 1 is x<= (-sqrt(57)-9)/2 \lor-8<= x<=-1\lor x>= (sqrt(57)-9)/2