What is |x^2-3x|+|x+1|>= 4 ?

The solution to |x^2-3x|+|x+1|>= 4 is x = 1

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|x^2-3x|+|x+1|>= 4

x^{2} - 3 x + ∣ x + 1 ∣ \geq 4

x \leq - 1 or x \geq 1

Solution steps

x^{2} - 3 x + ∣ x + 1 ∣ \geq 4

Find positive and negative intervals

x < - 1, - 1 \leq x \leq 0, 0 < x < 3, x \geq 3

Solve the inequality for each interval

x < - 1 or x = - 1 or 1 \leq x < 3 or x \geq 3

Merge Overlapping Intervals

x \leq - 1 or x \geq 1

\frac{( x - 4 )}{( x + 2 )} \leq \frac{( x - 2 )}{( x - 1 )}

x^{2} + x - 6 < 6

∣ x - 1 ∣ \geq ∣ x - 3 ∣

(1 + x)^{2} \geq 1 - x^{2}

x^{2} - 6 x + 8 \leq 4 - x

What is |x^2-3x|+|x+1|>= 4 ?
The solution to |x^2-3x|+|x+1|>= 4 is x<=-1\lor x>= 1