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Math Help - quadratic equation w/absoulute values

  1. #1
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    quadratic equation w/absoulute values

    how would you solve
    x^2+|2x-1|=3
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  2. #2
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    Quote Originally Posted by requal View Post
    how would you solve
    x^2+|2x-1|=3
    interesting ...

    |2x-1| = 3-x^2

    since |2x-1| \geq 0 , -\sqrt{3} \leq x \leq \sqrt{3}

    case 1 ...

    2x-1 = 3-x^2

    x^2+2x-4 = 0

    x^2+ 2x + 1 - 5 = 0

    (x + 1)^2 - 5 = 0

    x+1 = \pm \sqrt{5}

    x = -1 \pm \sqrt{5}

    one solution works ... x = -1 + \sqrt{5}

    case 2 ...

    2x-1 = -(3 - x^2)

    2x-1 = x^2 - 3

    0 = x^2 - 2x - 2

    0 = x^2 - 2x + 1 - 3

    0 = (x - 1)^2 - 3

    x-1 = \pm \sqrt{3}

    x = 1 \pm \sqrt{3}

    one solution works ... x = 1 - \sqrt{3}
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by requal View Post
    how would you solve
    x^2+|2x-1|=3
    recall, |x| = \left \{ \begin{array}{lr} x & \text{ if } x \ge 0 \\ & \\-x & \text{ if } x < 0 \end{array} \right.

    thus,

    x^2 + |2x - 1| = 3

    \Rightarrow |2x - 1| = 3 - x^2

    \Rightarrow 2x - 1 = 3 - x^2 ...........(if 2x - 1 \ge 0)

    or

    -(2x - 1) = 3 - x^2 ...........(if 2x - 1 < 0)

    now solve each equation independently, and then take the interval that covers both solutions
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