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

The solution to (2x^2-x-1)(-x+1)>= 0 is x<=-1/2 \lor x=1

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

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

x \leq - \frac{1}{2} or x = 1

Interval Notation

(- \infty, - \frac{1}{2}] \cup x = 1

Decimal Notation

x \leq - 0.5 or x = 1

Solution steps

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

Factor

(2 x^{2} - x - 1) (- x + 1) : - (x - 1)^{2} (2 x + 1)

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

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1)^{2} (2 x + 1)) (- 1) \leq 0 \cdot (- 1)

Simplify

(x - 1)^{2} (2 x + 1) \leq 0

Identify the intervals

x \leq - \frac{1}{2} or x = 1

∣ 5 x - 3 ∣ \geq 17

0.5

8 - 2 (a + 1) = 9 + a

- 6 (3 x - 2) = 4

5 yz + 10 z^{2}

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