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: a^{2} + a^{2} = (a + 1)^{2} a = ?
[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.
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.
solved it now a^{2} + a^{2} = a^{2} + 2a + 1 simplified it to a quadraticequation 0 = -a^{2} + 2a + 1 a_{1} = 1 + sq.root of 2 a_{2} = 1 - sq.root of 2 a_{2} is negative therefore a_{1} is the right answer which gives a = 2,41
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 | |.
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.