Which of the following represents the solution of the inequality 3x2 + 2x - 1 ≤ 0?

To solve the inequality 3x^2 + 2x - 1 ≤ 0, we first need to find the roots of the equation 3x^2 + 2x - 1 = 0.

Step 1: Find the roots of the equation We can use the quadratic formula to find the roots of the equation. The quadratic formula is given by:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

where a, b, and c are the coefficients of the quadratic equation. In this case, a = 3, b = 2, and c = -1.

Substituting these values into the quadratic formula gives:

x = [-2 ± sqrt((2)^2 - 43(-1))] / (2*3) x = [-2 ± sqrt(4 + 12)] / 6 x = [-2 ± sqrt(16)] / 6 x = [-2 ± 4] / 6

This gives us two roots: x = 1/3 and x = -2.

Step 2: Test the intervals Now we need to test the intervals determined by these roots in the original inequality. The intervals are (-∞, -2), (-2, 1/3), and (1/3, ∞).

Choose a test point in each interval and substitute it into the inequality. If the inequality is true, then that interval is part of the solution.

For (-∞, -2), choose x = -3. Substituting x = -3 into the inequality gives 3*(-3)^2 + 2*(-3) - 1 = 18 - 6 - 1 = 11, which is greater than 0. So, (-∞, -2) is not part of the solution.

For (-2, 1/3), choose x = 0. Substituting x = 0 into the inequality gives 3*(0)^2 + 2*0 - 1 = -1, which is less than 0. So, (-2, 1/3) is part of the solution.

For (1/3, ∞), choose x = 1. Substituting x = 1 into the inequality gives 3*(1)^2 + 2*1 - 1 = 3 + 2 - 1 = 4, which is greater than 0. So, (1/3, ∞) is not part of the solution.

Therefore, the solution to the inequality 3x^2 + 2x - 1 ≤ 0 is -2 ≤ x ≤ 1/3.