-2x^2+x>= 1 — step-by-step solution

Let's solve in

ℝ

- 2 x^{2} + x \geq 1

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

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = -2

b = 1

and

c = -1

Δ = 1 - 4 \cdot (-2) \cdot (-1)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-4 \cdot (-2)

8

■Multiplying by -1 gives the opposite:

8 \cdot (-1)

-8

Δ = 1 - 8

ⓘ Start from

1

and move back

8

on the number line, going below zero:

1 - 8

-7

Δ = -7

The quadratic has no real root: it has the sign of a =

-2

everywhere, so it is never positive.

▶Solution:

x \in \emptyset