Find the zeros of y = x2 – 6x – 4 by completing the square.

Question

Find the zeros of y = x2 – 6x – 4 by completing the square.

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Mộc Miên 1 month 2021-08-13T20:35:45+00:00 1 Answers 2 views 0

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    2021-08-13T20:37:43+00:00

    Answer:

    The solutions to the quadratic equations are:

    x=\sqrt{13}+3,\:x=-\sqrt{13}+3

    Step-by-step explanation:

    Given the function

    y\:=\:x^2\:-\:6x\:-\:4

    substitute y = 0 in the equation to determine the zeros

    0\:=\:x^2\:-\:6x\:-\:4

    Switch sides

    x^2-6x-4=0

    Add 4 to both sides

    x^2-6x-4+4=0+4

    Simplify

    x^2-6x=4

    Rewrite in the form (x+a)² = b

    But, in order to rewrite in the form x²+2ax+a²

    Solve for ‘a’

    2ax = -6x

    a = -3

    so add a² = (-3)² to both sides

    x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2

    x^2-6x+\left(-3\right)^2=13

    Apply perfect square formula:  (a-b)² = a²-2ab+b²

    \left(x-3\right)^2=13

    \mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}

    solve

    x-3=\sqrt{13}

    Add 3 to both sides

    x-3+3=\sqrt{13}+3

    Simplify

    x=\sqrt{13}+3

    now solving

    x-3=-\sqrt{13}

    Add 3 to both sides

    x-3+3=-\sqrt{13}+3

    Simplify

    x=-\sqrt{13}+3

    Thus, the solutions to the quadratic equations are:

    x=\sqrt{13}+3,\:x=-\sqrt{13}+3

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