Final answer:
The zeros of each quadratic equation are found using the quadratic formula or by factoring, leading to the zeros: A. ±2, B. two real solutions, C. ±3/2, and D. -2, -3.
Step-by-step explanation:
The zeros of a quadratic equation ax²+bx+c = 0 can be found using the quadratic formula, which is x = (-b ± √(b²-4ac))/(2a). For the given equations:
- For A. x² - 4 = 0, the zeros are x = ±2 because x² = 4.
- For B. 2x² + 3x + 1 = 0, we apply the quadratic formula to find the zeros: x = (-3 ± √(3²-4*2*1))/(2*2), which simplifies to two real solutions.
- For C. 4x² - 9 = 0, this is a difference of squares which can be written as (2x + 3)(2x - 3) = 0, giving zeros at x = ±3/2.
- For D. x² + 5x + 6 = 0, this can be factored to (x + 2)(x + 3) = 0, yielding zeros at x = -2 and x = -3.
For the additional equation x² + 1.2 x 10⁻²x - 6.0 × 10⁻³ = 0, using the quadratic formula will provide its zeros. The same method applies for x² +0.0211x -0.0211 = 0, rearranging it into standard quadratic form and then applying the quadratic formula.