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Simple solution?

  1. Aug 20, 2008 #1
    just looking for a quick solution for my equation, seems like my head is just working the wrong way coz I know it's not a hard one:

    a2 + a2 = (a + 1)2

    a = ?
     
  2. jcsd
  3. Aug 20, 2008 #2

    rock.freak667

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    Homework Helper

    a2+a2=2a2
    expand the right side and then simplify.
     
  4. Aug 20, 2008 #3
    i don't believe that (a + 1)(a + 1) is 2a2
    shouldn't that give a2 + 2a +1 ??
     
  5. Aug 20, 2008 #4
    [tex] a^{2} + a^{2} = (a+1)^{2} [/tex] simplifies to [tex] a^{2} + a^{2} = a^{2} + 2a + 1 [/tex] which when you move everything over to one side becomes [tex] a^{2} - 2a - 1 = 0[/tex] which is easy enough to solve. Not sure how rock.freak got what he did.
     
    Last edited: Aug 20, 2008
  6. Aug 20, 2008 #5
    I'm sure he was simplifying the left side (how much simpler can it be?). Then he said expand the RHS and rearrange to solve.
     
  7. Aug 20, 2008 #6
    solved it now

    a2 + a2 = a2 + 2a + 1

    simplified it to a quadraticequation

    0 = -a2 + 2a + 1

    a1 = 1 + sq.root of 2
    a2 = 1 - sq.root of 2

    a2 is negative therefore a1 is the right answer

    which gives a = 2,41
     
  8. Aug 20, 2008 #7
    Why can't a be negative?

    On the LHS you had [tex] a^{2} + a^{2} = (1 - \sqrt{2})^{2} + (1 - \sqrt{2})^{2} = 1 - 2 \sqrt{2} + 2 + 1 - 2 \sqrt{2} + 2 = 6 - 4 \sqrt{2} [/tex]

    However on the RHS you had [tex] (a+1)^{2} = (1 - \sqrt{2} + 1)^{2} = (2 - \sqrt{2})^{2} = 4 - 4 \sqrt{2} + 2 = 6 - 4 \sqrt{2} [/tex]

    Note also that the "simpler" way to do this would be to rewrite it as

    [tex] 2a^{2} = (a+1)^{2} \Rightarrow \sqrt{2} |a| = |a + 1| [/tex] and examine the appropriate regions to get rid of | |.
     
  9. Aug 20, 2008 #8
    The reason why it cannot be negative is that the origin of the problem was to determine the length of all sides of a likesided triangle thus can't be negative.

    I still think that using the quadratic equation is the simplest way of solving it.
     
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