Final answer:
After solving each quadratic equation through factoring or the quadratic formula, the solutions for the equations are paired with the solution sets as follows: A with {-4, 4}, B with {-5, 5}, C with {-8, 8}, and D with {-11, 11}.
Step-by-step explanation:
To match each quadratic equation with its solution set, we will solve each equation by factoring or using the quadratic formula, which is x = (-b ± √(b²-4ac)) / (2a) for equations of the form ax² + bx + c = 0. Below are the solutions for each given equation:
A. 2x²-32 = 0
To solve this, we factor out a '2' to get x²-16 = 0, which can be factored further as (x-4)(x+4) = 0. The solutions are x = 4 and x = -4, so the set is {-4, 4}.
B. 4x² - 100 = 0
To solve this, we factor as (2x-10)(2x+10) = 0. The solutions are x = 5 and x = -5, so the set is {-5, 5}.
C. x² - 55 = 9
First, we subtract 9 from both sides to get x² - 64 = 0, which can be factored as (x-8)(x+8) = 0. The solutions are x = 8 and x = -8, so the set is {-8, 8}.
D. x² - 140 = -19
By adding 19 to both sides, we get x² = 121, which implies x = 11 and x = -11, giving us the set {-11, 11}.
E. 2x² - 18=0
We divide the entire equation by 2 to get x² - 9 = 0, which factors into (x-3)(x+3) = 0. The solutions are x = 3 and x = -3, which can be written in the set {-3, 3}, but this set is not provided.
Hence, our pairings are A with 2, B with 3, C with 1, and D with 4. The equation E does not match with the given solution sets.