Find the number of real zeros of P(x) = (x-1)2+1:
(A) 1
(B) O
V o 2
(C) 2.
(D) None of these​

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

19.side- angle- side

20. angle- side- angle

step-by-step explanation: