\( 1 \leftarrow \) Solve each equation and inequality. Use set-builder or interval notation to write solution sets to the inequalities. (a) \( 2 x^{2}+5 x=0 \) (b) \( 2 x^{2}+5 x \leq 0 \) (c) \( 2 x^{2}+5 x \geq 0 \) A. The solution set is a list of numbers, (Simplify your answer. Type an integer or a fraction. Use a comma to separate answers as needed.) B. The solution set is an interval, (Simplify your answer. Type your answer in interval notation. Use integers or fractions for any numbers in the expression.) (a) \( 2 x^{2}+5 x=0 \). Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A

To solve the equation \( 2x^2 + 5x = 0 \), we can factor out \( x \): \( x(2x + 5) = 0 \) This gives us two solutions: 1. \( x = 0 \) 2. \( 2x + 5 = 0 \) → \( x = -\frac{5}{2} \) So, in set-builder notation, the solution set is: A. \( \{ 0, -\frac{5}{2} \} \) Now for the inequality \( 2x^2 + 5x \leq 0 \), we look at the critical points \( x = 0 \) and \( x = -\frac{5}{2} \). To find the intervals, we test values from the intervals formed by these points: 1. When \( x < -\frac{5}{2} \), say at \( x = -3 \): \( 2(-3)^2 + 5(-3) = 18 - 15 = 3 \) (not included). 2. Between \( -\frac{5}{2} \) and \( 0 \), say at \( x = -2 \): \( 2(-2)^2 + 5(-2) = 8 - 10 = -2 \) (included). 3. When \( x > 0 \), say at \( x = 1 \): \( 2(1)^2 + 5(1) = 2 + 5 = 7 \) (not included). Thus, the solution set for the inequality \( 2x^2 + 5x \leq 0 \) is: B. \( [-\frac{5}{2}, 0] \) Now for the inequality \( 2x^2 + 5x \geq 0 \), we consider the same critical points: 1. When \( x < -\frac{5}{2} \): Positive (not included). 2. Between \( -\frac{5}{2} \) and \( 0 \): Negative (not included). 3. When \( x > 0 \): Positive (included). Thus, the solution set for the inequality \( 2x^2 + 5x \geq 0 \) is: B. \( (-\infty, -\frac{5}{2}) \cup [0, \infty) \)