Answer: the equations are
x² + 2x - 15 = 0
x² + 2x + 20 = 35
(x + 5)(x - 3) = 0
Explanation:
To determine the quadratic equations that has the solution set {3,-5}, we would substitute x = 3 or
x = - 5 into each equation.
1) x² + 2x - 15 = 0
For x = 3,
3² + 2 × 3 - 15 = 0
9 + 6 - 15 = 0
0 = 0
For x = - 5,
- 5² + 2 × - 5 - 15 = 0
25 - 10 - 15 = 0
0 = 0
(3, 5) is a solution
2) x² - 2x + 10 = - 5
For x = 3,
3² - 2 × 3 + 10 = - 5
9 - 6 + 10 = - 5
13 = - 5
It is not equal equal. So, (3, - 5) is not a solution
3) x² + 2x + 20 = 35
For x = 3
3² + 2 × 3 + 20 = 35
9 + 6 + 20 = 35
35 = 35
For x = - 5
- 5² + 2 × - 5 + 20 = 35
25 - 10 + 20 = 35
35 = 35
(3, 5) is a solution
4) (x - 5)(x + 3) = 0
For x = 3
(3 - 5)(3 + 3) = 0
- 2 × 6 = 0
- 12 = 0
It is not equal equal. So, (3, - 5) is not a solution.
5) (x + 5)(x - 3) = 0
For x = 3
(3 + 5)(3 - 3) = 0
0 = 0
For x = - 5,
(- 5 + 5)(- 5 - 3) = 0
0 = 0
(3, 5) is a solution