What is the solution set of the quadratic inequality 2x^2 + 4x + 2 ≥ 18?

Question

What is the solution set of the quadratic inequality 2x^2 + 4x + 2 ≥ 18? ○ x ∈ [-4, -1) ∪ (-1, 2] ○ x ∈ (-∞, -4] ∪ [2, ∞) ○ x ∈ [-4, 2] ○ x ∈ (-4, 2)

SnapAndSolve AI · Accepted Answer

To solve the inequality 2x^2 + 4x + 2 ≥ 18, we first rewrite it as 2x^2 + 4x + 2 - 18 ≥ 0, which simplifies to 2x^2 + 4x - 16 ≥ 0. Dividing the entire inequality by 2 gives x^2 + 2x - 8 ≥ 0.

Next, we find the roots of the equation x^2 + 2x - 8 = 0 using the quadratic formula: x = [-b ± √(b² - 4ac)] / 2a, where a = 1, b = 2, and c = -8.

Calculating the discriminant: b² - 4ac = 2² - 4(1)(-8) = 4 + 32 = 36.

So, the roots are x = [-2 ± √36] / 2 = [-2 ± 6] / 2, which gives x = 2 and x = -4.

The inequality x^2 + 2x - 8 ≥ 0 is satisfied for x ∈ (-∞, -4] ∪ [2, ∞).

x ∈ [-4, -1) ∪ (-1, 2] - Incorrect
x ∈ (-∞, -4] ∪ [2, ∞) - Correct
x ∈ [-4, 2] - Incorrect
x ∈ (-4, 2) - Incorrect

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What is the solution set of the quadratic inequality 2x^2 + 4x + 2 ≥ 18?

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