What is x-2>= (x-2)(2x^2-x-15) ?

The solution to x-2>= (x-2)(2x^2-x-15) is x<= (1-sqrt(129))/4 \lor 2<= x<= (1+sqrt(129))/4

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x-2>= (x-2)(2x^2-x-15)

x - 2 \geq (x - 2) (2 x^{2} - x - 15)

x \leq \frac{1 - 129}{4} or 2 \leq x \leq \frac{1 + 129}{4}

Interval Notation

(- \infty, \frac{1 - 129}{4}] \cup [2, \frac{1 + 129}{4}]

Decimal Notation

x \leq - 2.58945 \dots or 2 \leq x \leq 3.08945 \dots

Solution steps

x - 2 \geq (x - 2) (2 x^{2} - x - 15)

Rewrite in standard form

14 x - 32 + 5 x^{2} - 2 x^{3} \geq 0

Factor

14 x - 32 + 5 x^{2} - 2 x^{3} : - (x - 2) (x - \frac{1 + 129}{4}) (x - \frac{1 - 129}{4})

- (x - 2) (x - \frac{1 + 129}{4}) (x - \frac{1 - 129}{4}) \geq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 2) (x - \frac{1 + 129}{4}) (x - \frac{1 - 129}{4})) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 2) (x - \frac{1 + 129}{4}) (x - \frac{1 - 129}{4}) \leq 0

Identify the intervals

x \leq \frac{1 - 129}{4} or 2 \leq x \leq \frac{1 + 129}{4}

(\frac{15}{14})^{2} + (\frac{15}{14})^{2} = c^{2}

4 x^{2} = - 2 x + 12

∣ x + 6 ∣ \geq 2

- 1 - \frac{1}{3}

3 x^{2} + 2 x = 5

What is x-2>= (x-2)(2x^2-x-15) ?
The solution to x-2>= (x-2)(2x^2-x-15) is x<= (1-sqrt(129))/4 \lor 2<= x<= (1+sqrt(129))/4