What is |x-3|+|x+4|>= 12 ?

The solution to |x-3|+|x+4|>= 12 is x = 11/2

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|x-3|+|x+4|>= 12

∣ x - 3 ∣ + ∣ x + 4 ∣ \geq 12

x \leq - \frac{13}{2} or x \geq \frac{11}{2}

Solution steps

∣ x - 3 ∣ + ∣ x + 4 ∣ \geq 12

Find positive and negative intervals

x < - 4, - 4 \leq x < 3, x \geq 3

Solve the inequality for each interval

x \leq - \frac{13}{2} or No Solution or x \geq \frac{11}{2}

Merge Overlapping Intervals

x \leq - \frac{13}{2} or x \geq \frac{11}{2}

x^{2} - 2 x + ∣ x - 4 ∣ > x^{2} - 3 x + 4

x^{2} - 4 x < 5

x^{2} - 5 x < 6

x^{2} - 2 x < x

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

What is |x-3|+|x+4|>= 12 ?
The solution to |x-3|+|x+4|>= 12 is x<=-13/2 \lor x>= 11/2