What is |x^2-4x|<= |x| ?

The solution to |x^2-4x|<= |x| is x=0\lor 3<= x<= 5

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|x^2-4x|<= |x|

x^{2} - 4 x \leq ∣ x ∣

x = 0 or 3 \leq x \leq 5

Solution steps

x^{2} - 4 x \leq ∣ x ∣

Find positive and negative intervals

x \leq 0, 0 < x < 4, x \geq 4

Solve the inequality for each interval

x = 0 or 3 \leq x < 4 or 4 \leq x \leq 5

Merge Overlapping Intervals

x = 0 or 3 \leq x \leq 5

\frac{2 ( 4 p + 1 )}{5} \leq \frac{3 + 12 p}{4}

224 a z + 56 a c - 84 yz - 21 yc

\frac{- 6 + 8 i}{- 4 - 3 i}

\frac{8 + 5 i}{3 - 2 i}

9 \leq 4 x - 11 \leq 21