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Thread: quadratic equation w/absoulute values

  1. #1
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    quadratic equation w/absoulute values

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

    $\displaystyle |2x-1| = 3-x^2$

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

    case 1 ...

    $\displaystyle 2x-1 = 3-x^2$

    $\displaystyle x^2+2x-4 = 0$

    $\displaystyle x^2+ 2x + 1 - 5 = 0$

    $\displaystyle (x + 1)^2 - 5 = 0$

    $\displaystyle x+1 = \pm \sqrt{5}$

    $\displaystyle x = -1 \pm \sqrt{5}$

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

    case 2 ...

    $\displaystyle 2x-1 = -(3 - x^2)$

    $\displaystyle 2x-1 = x^2 - 3$

    $\displaystyle 0 = x^2 - 2x - 2$

    $\displaystyle 0 = x^2 - 2x + 1 - 3$

    $\displaystyle 0 = (x - 1)^2 - 3$

    $\displaystyle x-1 = \pm \sqrt{3}$

    $\displaystyle x = 1 \pm \sqrt{3}$

    one solution works ... $\displaystyle x = 1 - \sqrt{3}$
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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
    $\displaystyle x^2+|2x-1|=3$
    recall, $\displaystyle |x| = \left \{ \begin{array}{lr} x & \text{ if } x \ge 0 \\ & \\-x & \text{ if } x < 0 \end{array} \right.$

    thus,

    $\displaystyle x^2 + |2x - 1| = 3$

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

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

    or

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

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