Question

Find all values of n for which the equation has two complex (non-real) solutions.

(n + 1)r² + 3r + 9 = 0
Write your answer starting with n, followed by an equals sign or inequality symbol (for
example, n < 5). Reduce all fractions.

Answer 1

n > -3/4

Explanation:

In order for the solutions to be complex, the discriminant must be negative. In other words, b²-4ac<0.

a=n+1

b=3

c=9

3²-4(9)(n+1) < 0

-36n < 27

n > -3/4

Check... n = -2: a=-1, b=3, c=9 ⇒ 3²-4(-2)(9) = 9+72 = 81 (positive)

n = 0: a=1, b=3, c=9 ⇒ 3²-4(1)(9) = 9-36 = -27 (negative)