Question

A) 4x^2 + 4x + 36 = 0

A) x = -3, x = -6
B) x = -2, x = -8
C) x = -6, x = -10
D) x = -4, x = -9

b) 3x^2 - 18x - 48 = 0

A) x = 4, x = -4
B) x = -4, x = 4
C) x = -2, x = 12
D) x = 6, x = -8

Answer 1

None of the provided options for the first quadratic equation are correct, as it does not have real solutions. For the second equation, the solutions are x = 6 and x = -8, matching option D.

Step-by-step explanation:

The student has provided two separate quadratic equations to be solved. For any quadratic equation of the form ax² + bx + c = 0, the solutions or roots can be found using the quadratic formula, which is x = [-b ± √(b² - 4ac)] / (2a).

Let's solve each equation separately.

a) 4x² + 4x + 36 = 0.
Here, a = 4, b = 4, and c = 36. Applying the quadratic formula does not yield any real solutions because the discriminant (b² - 4ac) is negative. Hence, none of the given options A, B, C, or D are correct.

b) 3x² - 18x - 48 = 0.
Here, a = 3, b = -18, and c = -48. Substituting these values into the quadratic formula, we find the solutions to be x = 6 and x = -8, which corresponds to option D.