What is 4x^3-7x^2>= 2x ?

The solution to 4x^3-7x^2>= 2x is -1/4 = 2

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4x^3-7x^2>= 2x

4 x^{3} - 7 x^{2} \geq 2 x

- \frac{1}{4} \leq x \leq 0 or x \geq 2

Interval Notation

[- \frac{1}{4}, 0] \cup [2, \infty)

Decimal Notation

- 0.25 \leq x \leq 0 or x \geq 2

Solution steps

4 x^{3} - 7 x^{2} \geq 2 x

Rewrite in standard form

4 x^{3} - 7 x^{2} - 2 x \geq 0

Factor

4 x^{3} - 7 x^{2} - 2 x : x (4 x + 1) (x - 2)

x (4 x + 1) (x - 2) \geq 0

Identify the intervals

- \frac{1}{4} \leq x \leq 0 or x \geq 2

(- \frac{1}{x})^{4}

9 a^{2} b^{- 4}

7 x^{2} + 4 x - 2 = 0

- (\frac{3}{2} + \frac{5}{2} i) + (\frac{5}{3} + \frac{11}{3} i)

- 2 x^{3} - 6 x^{2} + 56 x

What is 4x^3-7x^2>= 2x ?
The solution to 4x^3-7x^2>= 2x is -1/4 <= x<= 0\lor x>= 2