solve for x to the nearest tenth y completing the square: x^2-5x+7=0​

Question

solve for x to the nearest tenth y completing the square: x^2-5x+7=0​

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Adela 1 year 2021-09-03T20:55:15+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-09-03T20:56:49+00:00

    Answer:

    x = (1/2)(5 ± i√3)

    Step-by-step explanation:

    x² – 5x + 7 = 0

    subtract 7 from both sides

    x² – 5x = -7

    Use half the x coefficent, -5/2, as the complete the square term

    (x – 5/2)² = -7 + (-5/2)²

    (x – 5/2)² = -7 + 25/4

    (x – 5/2)² = -3/4

    Take the square root of both sides

    x – 5/2 = ±(√-3) / 2

    x – 5/2 = ±(i√3) / 2

    Add 5/2 to both sides

    x = 5/2 ± (i√3) / 2

    factor out 1/2

    x = (1/2)(5 ± i√3)

    0
    2021-09-03T20:57:13+00:00

    Answer:

    does not have a solution because √-0.75 ≠ R

    Step-by-step explanation:

    x^2 – 5x + (5/2)^2 = -7

    x^2 – 5x + 6.25 = -7 + 6.25

    (x – 2.5)^2 = -0.75

    (x – 2.5) = √-0.75

    does not have a solution because √-0.75 ≠ R

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