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

The solution to |x^2+x-5|<= x-1 is sqrt(7)-1<= x<= 2

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

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

7 - 1 \leq x \leq 2

Interval Notation

[7 - 1, 2]

Decimal Notation

1.64575 \dots \leq x \leq 2

Solution steps

x^{2} + x - 5 \leq 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

No Solution or 7 - 1 \leq x < \frac{21 - 1}{2} or \frac{21 - 1}{2} \leq x \leq 2

Merge Overlapping Intervals

7 - 1 \leq x \leq 2

\frac{1 5 ^{2}}{5}

\frac{- 1}{2} x \leq 17

36

x + \frac{1}{5} x = 36

3 n^{2} - n - 10