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algebra

posted by on .

solve for w: the square root of 2-w=w


2-w=w^
w^+w-2
(w+2) (w-10)
w+2=0 w-1=0
w=-2 w=1

  • algebra - ,

    √(2-w) = w^2
    2-w = w^4
    w^4+2w-2 = 0
    The solutions to that are not trivial.

    So, assuming you meant

    √2 - w = w
    √2 = 2w
    2 = 4w^2
    w = ±1/√2

    Better try again. Ignoring the first two lines, once you get to

    (w+2)(w-10)
    I'm sure you meant
    (w+2)(w-1) = 0
    In which case the roots are indeed -2 and 1

    You seriously need to improve you exponential notation!

  • algebra - ,

    Wow, that is some creative algebra

    I assume your equation is
    √(2-w) = w , or is it √2 - w = w
    (notice how my brackets totally change the meaning of the problem)

    I will assume the first
    square both sides

    (√(2-w) )^2 = w^2
    2 - w = w^2
    w^2 + w - 2 = 0
    (w+2)(w-1) = 0
    w = -2 or w = 1

    Since we squared, both solutions MUST be verified
    if x = 1
    LS = √(2-1) = 1
    RS = 1^2 = 1 = LS
    so x = 1

    if x = -2
    LS = √(2 + 2) = √4 = 2
    RS = -2
    LS ≠ RS

    so the only solution is x = 1

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