What is |x^2+x|>|x^2+5x-8| ?

The solution to |x^2+x|>|x^2+5x-8| is x<-4\lor 1<x<2

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|x^2+x|>|x^2+5x-8|

x^{2} + x > x^{2} + 5 x - 8

x < - 4 or 1 < x < 2

Solution steps

x^{2} + x > x^{2} + 5 x - 8

Find positive and negative intervals

x \leq \frac{- 57 - 5}{2}, \frac{- 57 - 5}{2} < x \leq - 1, - 1 < x < 0, 0 \leq x < \frac{57 - 5}{2}, x \geq \frac{57 - 5}{2}

Solve the inequality for each interval

x \leq \frac{- 57 - 5}{2} or \frac{- 57 - 5}{2} < x < - 4 or No Solution or 1 < x < \frac{57 - 5}{2} or \frac{57 - 5}{2} \leq x < 2

Merge Overlapping Intervals

x < - 4 or 1 < x < 2

∣ x + 3 ∣ > ∣ 2 x - 1 ∣

x^{2} + x - 5 < 0

∣ x - 2 ∣ > ∣ 2 x - 3 ∣

x^{2} - 7 x + 12 < ∣ x - 4 ∣

x^{2} + 3 x > 2 - x^{2}

What is |x^2+x|>|x^2+5x-8| ?
The solution to |x^2+x|>|x^2+5x-8| is x<-4\lor 1<x<2