What is -64x(-16x^2+1)>= 0 ?

The solution to -64x(-16x^2+1)>= 0 is -1/4 = 1/4

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-64x(-16x^2+1)>= 0

- 64 x (- 16 x^{2} + 1) \geq 0

- \frac{1}{4} \leq x \leq 0 or x \geq \frac{1}{4}

Solution steps

- 64 x (- 16 x^{2} + 1) \geq 0

Rewrite in standard form

16 x^{3} - x \geq 0

Factor

16 x^{3} - x : x (4 x + 1) (4 x - 1)

x (4 x + 1) (4 x - 1) \geq 0

Identify the intervals

- \frac{1}{4} \leq x \leq 0 or x \geq \frac{1}{4}

s^{3} - 64 t^{3}

∣ 7 x + 1 ∣ - 1 = 11

- 5 z^{2} - 3 z - 11 = - 6 z^{2}

x^{2} = 128

x + 7 = 17

What is -64x(-16x^2+1)>= 0 ?
The solution to -64x(-16x^2+1)>= 0 is -1/4 <= x<= 0\lor x>= 1/4