What is 24x^4+2x^3-5x^2-x>= 0 ?

The solution to 24x^4+2x^3-5x^2-x>= 0 is x = 1/2

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24x^4+2x^3-5x^2-x>= 0

24 x^{4} + 2 x^{3} - 5 x^{2} - x \geq 0

x \leq - \frac{1}{3} or - \frac{1}{4} \leq x \leq 0 or x \geq \frac{1}{2}

Interval Notation

(- \infty, - \frac{1}{3}] \cup [- \frac{1}{4}, 0] \cup [\frac{1}{2}, \infty)

Decimal Notation

x \leq - 0.33333 \dots or - 0.25 \leq x \leq 0 or x \geq 0.5

Solution steps

24 x^{4} + 2 x^{3} - 5 x^{2} - x \geq 0

Factor

24 x^{4} + 2 x^{3} - 5 x^{2} - x : x (2 x - 1) (4 x + 1) (3 x + 1)

x (2 x - 1) (4 x + 1) (3 x + 1) \geq 0

Identify the intervals

x \leq - \frac{1}{3} or - \frac{1}{4} \leq x \leq 0 or x \geq \frac{1}{2}

4 x^{3} - 9 x^{2} - 11 x + 26 \geq 0

x^{2} + x + 3 < 0

\frac{1}{( 4 x - 1 )} > 2

(x - 4) (x + 5) > 0

\frac{1}{( 2 x + 1 )} > 0

What is 24x^4+2x^3-5x^2-x>= 0 ?
The solution to 24x^4+2x^3-5x^2-x>= 0 is x<=-1/3 \lor-1/4 <= x<= 0\lor x>= 1/2