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​

Question

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​

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Delwyn 4 years 2021-09-03T19:23:46+00:00 1 Answers 15 views 0

Answers ( )

  1. Nem
    0
    2021-09-03T19:25:44+00:00
    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

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )