What is |x^2+x-5|>= x-1 ?

The solution to |x^2+x-5|>= x-1 is x = 2

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|x^2+x-5|>= x-1

x^{2} + x - 5 \geq x - 1

x \leq 7 - 1 or x \geq 2

Interval Notation

(- \infty, 7 - 1] \cup [2, \infty)

Decimal Notation

x \leq 1.64575 \dots or x \geq 2

Solution steps

x^{2} + x - 5 \geq x - 1

Find positive and negative intervals

x \leq \frac{- 21 - 1}{2}, \frac{- 21 - 1}{2} < x < \frac{21 - 1}{2}, x \geq \frac{21 - 1}{2}

Solve the inequality for each interval

x \leq \frac{- 21 - 1}{2} or \frac{- 21 - 1}{2} < x \leq 7 - 1 or x \geq 2

Merge Overlapping Intervals

x \leq 7 - 1 or x \geq 2

- \frac{1}{3} + \frac{7}{5} - \frac{8}{6}

7 y - 6 y - 10 = 13

y - 8 = 0

x^{2} = \frac{36}{121}

\frac{5 + 7 i}{5 + 2 i}

What is |x^2+x-5|>= x-1 ?
The solution to |x^2+x-5|>= x-1 is x<= sqrt(7)-1\lor x>= 2