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solve 2y=7sqrt(y)-3

  • math -

    2y=7sqrt(y)-3
    2y+3 = √y
    square both sides
    4y^2 + 12y + 9 = y
    4y^2 + 11y + 9 = 0
    y = (-11 ±√-23)/8

    I am assuming you are dealing with real numbers only
    since y is a complex number, and we cannot take a square root of a complex number, your equation has no real solution.

  • math -

    Using the substitution y=e^x,solve 2e^x=7√e^x -3

  • math -

    for some reason , I dropped the 7 on the right side

    let's start again
    square both sides

    4y^2 + 12y + 9 = 49y
    4y^2 - 37y + 9 = 0
    y = (37 ± √1225)/8
    y = 9 or y = 0.25

    (it would have factored to (y-9)(4y - 1)=0 )

    now now
    e^x = 9
    take ln of both sides
    ln (e^x) = ln9
    x = ln9

    or

    e^x = .25 or 1/4
    take ln again
    x = ln (1/4) = ln1 - ln4 = -ln4

    x = ln9 or x = -ln4

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