Simple solution?

  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. rock.freak667

    rock.freak667 6,230
    Homework Helper

    a2+a2=2a2
    expand the right side and then simplify.
     
  4. i don't believe that (a + 1)(a + 1) is 2a2
    shouldn't that give a2 + 2a +1 ??
     
  5. [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. 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. 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. 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. 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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