Answer: B Step-by-step explanation: Using the determinant to determine the type of zeros Given f(x) = ax² + bx + c ( a ≠ 0 ) ← in standard form, then the discriminant is Δ = b² – 4ac • If b² – 4ac > 0 then 2 real and distinct zeros • If b² – 4ac = 0 then 2 real and equal zeros • If b² – 4ac < 0 then 2 complex zeros Given f(x) = (x – 1)² + 1 ← expand factor and simplify = x² – 2x + 1 + 1 = x² – 2x + 2 ← in standard form with a = 1, b = – 2, c = 2, then b² – 4ac = (- 2)² – (4 × 1 × 2) = 4 – 8 = – 4 Since b² – 4ac < 0 then the zeros are complex Thus P(x) has no real zeros Log in to Reply

Answer:B

Step-by-step explanation:Using the determinant to determine the type of zeros

Given

f(x) = ax² + bx + c ( a ≠ 0 ) ← in standard form, then the discriminant is

Δ = b² – 4ac

• If b² – 4ac > 0 then 2 real and distinct zeros

• If b² – 4ac = 0 then 2 real and equal zeros

• If b² – 4ac < 0 then 2 complex zeros

Given

f(x) = (x – 1)² + 1 ← expand factor and simplify

= x² – 2x + 1 + 1

= x² – 2x + 2 ← in standard form

with a = 1, b = – 2, c = 2, then

b² – 4ac = (- 2)² – (4 × 1 × 2) = 4 – 8 = – 4

Since b² – 4ac < 0 then the zeros are complex

Thus P(x) has no real zeros