Math- Algebra II

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x^2-4x+12=0 Complete the square to solve for x.

Okay I know how to do up to here:

(-4/2)^2 -12=x^2-4x(-4/2)^2

2-12= x^2-4x+2

  • Math- Algebra II -

    x^2 - 4x = -12
    x^2 - 4x + 4 = -12 + 4
    (x-2)^2 = -8
    x-2 = ± √-8 , but √-8 = 2√2 √-1 = 2i√2

    x = 2 ± 2i√2

  • Math- Algebra II -

    Easist way always is to go and get the roots:

    (-b + sqrt(b^2 - 4ac)) / 2a
    x^2 -4x +12 = 0

    a= 1, b=-4 c = 12

    (-(-4) + sqrt(16 - 4*12)) / 2

    (4 + sqrt(-32))/2

    4 + sqrt(i^2 * 16 *2))/2

    note : i^2 = -1

    (4 + i 4 sqrt(2))/2


    2+ 2i sqrt(2)

    are the two roots

    Hope this helps Christine!

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