Answer:
The quadratic equation that has complex roots is the equation;
B, 9·x² + 33 = 0
Explanation:
The given quadratic equation are;
A. 2·(x + 3)² = 64
B. 9·x² + 33 = 0
C. (x, y)
(-2.5, 2), (-2, 0), (-1.5, -1), (-1, -1), (-0.5, 0), (0, 2)
D. A graph with x-intercept at (2, 0)
Analyzing each of the quadratic equations, we have;
For A
2·(x + 3)² = 64 gives;
(x + 3)² = 64/2 = 32
(x + 3)² = 32 = (4·√2)²
∴ (x + 3)² = (4·√2)²
(x + 3) = ±4·√2
x = 4·√2 - 3 or x = -4·√2 - 3 which are real numbered roots
For option B we have;
9·x² + 33 = 0
9·x² = -33
9·x² = (3·x)² = -33
(3·x)² = -33
3·x = ±√(-33)
x = ±√(-33)/3 = ± √(-1)×(√33)/3 which has complex root √(-1)
For option C we have;
The roots of the equation occurs at y = 0 which corresponds with the points (-2, 0) and (-0.5, 0), therefore, the equation has real roots
For option C we have;
The roots of the equation occurs at y = 0 which is found as the point (2, 0) therefore, the equation has real roots