Find the quadratic function with integral coefficients having the given zeros:?

Question

1. -6 and 0
2. -2, 7/2
3. 3/2, -1/2
4. 3+i and 3-i
5. 2+ √2 and 2- √2
6. 5+3i and 5-3i
pls with complete solution...

Answer 1

The method is pretty straightforward. Start with your solutions, multiply them out and reduce.


PROBLEM 1:
x = -6 or x = 0

Write them in the form (x - a)(x - b) = 0
(x - (-6))(x - 0) = 0

Simplify:
(x + 6)x = 0

Multiply through:
x² + 6x = 0


PROBLEM 2:
x = -2 or x = 7/2

Write in (x - a)(x - b) = 0 form:
(x - (-2))(x - 7/2) = 0

Simplify:
(x + 2)(x - 7/2) = 0

Multiply by the denominators to remove any fractions (i.e. multiply both sides by 2):
(x + 2) * 2 (x - 7/2) = 2 * 0
= (x + 2)(2x - 7) = 0

Multiply through:
2x² + 4x - 7x - 14 = 0

Combine terms:
2x² - 3x - 14 = 0


PROBLEM 3:
(x - 3/2)(x + 1/2) = 0

Multiply both parts by 2 to remove fractions:
2(x - 3/2) * 2(x + 1/2) = 0 * 2 * 2 = 0

Simplify:
(2x - 3)(2x + 1) = 0

Multiply out:
4x² - 6x + 2x - 3 = 0

Group:
4x² - 4x - 3 = 0


PROBLEM 4:
(x - (3+i))(x - (3-i)) = 0

Multiply out:
x² - (3+i)x - (3-i)x + (3+i)(3-i)

Simplify:
x² -3x - ix - 3x + ix + (3² + 3i - 3i - i²)

Cancel out -ix and ix, cancel +3i and -3i:
x² - 6x + (9 - i²)

Replace i² = -1:
x² - 6x + (9 - (-1))
x² - 6x + 10


PROBLEM 5:
(x - (2+√2))(x - (2-√2))

Multiply out:
x² - (2+√2)x - (2-√2)x + (2+√2)(2-√2)
x² - 2x - √2x - 2x + √2x + (2² + 2√2 - 2√2 - (√2)²)

Cancel -√2x and +√2x as well as 2√2 and -2√2:
x² - 4x + (4 - (√2)²)

Replace (√2)² = 2
x² - 4x + (4 - 2)
x² - 4x + 2


PROBLEM 6:
I've given you enough tips above, I think you can do this one on your own...