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Confused About Squareroots & Absolute Values

  1. Mar 13, 2013 #1
    $${ x }^{ 2 }=4\\ \sqrt { { x }^{ 2 } } =\sqrt { 4 } \\ |x|=2$$
    According to my professor, in the above case, the absolute value gives two solutions: ##x=±2##
    Consider the discriminant in the quadratic formula: $$x=\frac { -b±\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ Let\quad { z }^{ 2 }={ b }^{ 2 }-4ac\\ ±\sqrt { { b }^{ 2 }-4ac } \\ =±\sqrt { { z }^{ 2 } } \\ =±|z|\\ =±z$$
    However, according to my professor, in this case, the absolute value gives only one solution: ##|z|=z##

    How come the absolute value sometimes gives one solution and sometimes it gives two solutions?
     
    Last edited by a moderator: Mar 13, 2013
  2. jcsd
  3. Mar 13, 2013 #2

    Mark44

    Staff: Mentor

    This makes no sense to me. If z is ≥ 0, then |z| = z. OTOH, if z < 0, then |z| = -z.
     
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