What is |x-4|>= x^2-7x+12 ?

The solution to |x-4|>= x^2-7x+12 is 2<= x<= 4

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|x-4|>= x^2-7x+12

∣ x - 4 ∣ \geq x^{2} - 7 x + 12

2 \leq x \leq 4

Interval Notation

[2, 4]

Solution steps

∣ x - 4 ∣ \geq x^{2} - 7 x + 12

Find positive and negative intervals

x < 4, x \geq 4

Solve the inequality for each interval

2 \leq x < 4 or x = 4

Merge Overlapping Intervals

2 \leq x \leq 4

1 - (\frac{11}{61})^{2}

7 y^{2} - 6 y = 0

\frac{2 x + 3 y}{2 y ^{2} + 2 x - 8 y}

i^{4} + 4 i^{2} + 4

4 - 6 ∣ - 6 - 3 x ∣ \leq - 5