how to write a quadratic function that has zeros of 9 ± 7i

Answer 1

We find zeros when we set factors = 0

If the zero is 9 + 7i, then the factor is (x - 9 - 7i)

because if we set x = 9 + 7i, then x - 9 - 7i = 0

Similarly, if the zero is 9 - 7i, then the factor is (x - 9 + 7i)

because if we set x = 9 - 7i, then x - 9 + 7i = 0

Therefore our function is f(x) = (x - 9 - 7i)(x - 9 + 7i)

We can also write this as f(x) = (x - (9 + 7i))(x - (9 - 7i)) by factoring a negative out of the complex numbers.

If we need to simplify, we can now FOIL (First Outside Inside Last):

x² - (9 - 7i)*x - (9 + 7i)*x + -(9 + 7i)*-(9 - 7i)

x² - 9x + 7ix - 9x - 7ix + (9 + 7i)*(9 - 7i)

Since 7ix - 7ix = 0, we now have x² - 18x + (9 + 7i)*(9 - 7i)

If we FOIL the last term, we have:

(9 + 7i)*(9 - 7i)

81 - 63i + 63i - 49i²

Since i² = -1 and -63i + 63i = 0, we have 81 + 49 = 130

Thus, our equation is x² - 18x + 130

Answer 2

(x-9)^2 = -49

x^2 -18x + 81 = -49

x^2 -18x +130 = 0

x-9 = + or - 7i

x = 9 + or - 7i