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College Algebra

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This is my question 2x^2-8x=-25
I cant seem to figure out how to solve for x?

  • College Algebra -

    Add the -25 to the other side...
    2x^2-8x+25=0,
    Now use the quadratic formula to solve for x...
    Hope this helps

  • College Algebra -

    I was told to start with
    x^2-8x+16=7/2 so where do i go from there?

  • College Algebra -

    2x^2 - 8x =-25
    divide each term by 2

    using completing the square,

    x^2 - 4x = -25/2
    take 1/2 the coefficient of the x term, square it, then add it to each side

    x^2 - 4x +4 = -25/2 +4
    (x-2)^2 = -17/2 or -34/4 , (I anticipated to take √ of both sides)
    x -2 = ± √-34/2 = ± i√34/2

    x = 4/2 ± i√34/2 = (4 ± i√34)/2

    or ... by the formula

    2x^2 - 8x + 25 = 0
    x = (8 ± √-136)/4
    = (8 ± 2√-34)/2
    = (4 ± i√34)/2

  • College Algebra -

    I was told to start with
    x^2-8x+16=7/2 so where do i go from there? That's after / by 2.

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