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Math Help - Solving for x with square root

  1. #1
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    Solving for x with square root

    x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

    How do I solve for x?

    Trying to get the x on the same side, so x(x + 2) = -(1/2) + sqroot(1/4 - a) - a

    But can not get any further
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  2. #2
    Bar0n janvdl's Avatar
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    Square both sides of the equation.
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  3. #3
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    x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

    x^2 +2x = -a -1/2 +sqrt(1/4 -a)
    x^2 +2x +1 = 1 -a -1/2 +sqrt(1/4 -a)
    (x +1)^2 = 1/2 -a +sqrt(1/4 -a)

    I am not successful yet on how to make the RHS a perfect square too.
    So, for now,

    x +1 = +,-sqrt[1/2 -a +sqrt(1/4 -a)]
    x = -1 +,-sqrt[1/2 -a +sqrt(1/4 -a)] ------------answer.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tikitac View Post
    x^2 + 2x + a = -(1/2) + sqroot(1/4 - a)

    How do I solve for x?

    Trying to get the x on the same side, so x(x + 2) = -(1/2) + sqroot(1/4 - a) - a

    But can not get any further
    This is the same as ticbol's answer, but in a different way.

    a is a constant, so:
    x^2 + 2x + a = -\frac{1}{2} + \sqrt{ \frac{1}{4} - a}

    x^2 + 2x + \left ( \frac{1}{2} + a - \sqrt{ \frac{1}{4} - a} \right ) = 0

    This is of the form bx^2 + cx + d = 0, so we may use the quadratic formula:
    x = \frac{ -c \pm \sqrt{c^2 - 4bd} }{2b}
    (Sorry for the shift in variables, but you already have an "a" in your expression. I didn't want to confuse the symbols.)

    So
    x = \frac{-2 \pm \sqrt{4 - 4 \left ( \frac{1}{2} + a - \sqrt{ \frac{1}{4} - a} \right ) }}{2}

    x = \frac{-2 \pm 2\sqrt{1 - \frac{1}{2} - a + \sqrt{ \frac{1}{4} - a} }}{2}

    x = -1 \pm \sqrt{ \frac{1}{2} - a + \sqrt{ \frac{1}{4} - a} }

    -Dan
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