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What is the solution set of the quadratic inequality -2(x + 2)(x - 5) + 3 ≥ 3?

What is the solution set of the quadratic inequality -2(x + 2)(x - 5) + 3 ≥ 3?
○ x ∈ [-5, 2]
○ x ∈ (-2, 5)
○ x ∈ [-2, 5]
○ x ∈ (-∞, 2) ∪ [5, ∞)

Answer

First, simplify the inequality:

Start with: -2(x + 2)(x - 5) + 3 ≥ 3
Subtract 3 from both sides: -2(x + 2)(x - 5) ≥ 0
Expand: -2(x^2 - 5x + 2x - 10) ≥ 0
Simplify: -2(x^2 - 3x - 10) ≥ 0
Distribute: -2x^2 + 6x + 20 ≥ 0

Now, solve the inequality:

Factor the quadratic: -2(x + 2)(x - 5) ≥ 0
Find the critical points: x = -2 and x = 5
Test intervals: (-∞, -2), (-2, 5), (5, ∞)
For x ∈ (-∞, -2): Choose x = -3, -2(-3 + 2)(-3 - 5) = -2(-1)(-8) = 16 ≥ 0 (True)
For x ∈ (-2, 5): Choose x = 0, -2(0 + 2)(0 - 5) = -2(2)(-5) = 20 ≥ 0 (True)
For x ∈ (5, ∞): Choose x = 6, -2(6 + 2)(6 - 5) = -2(8)(1) = -16 ≥ 0 (False)

The solution set is x ∈ (-∞, -2) ∪ [5, ∞).

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