simple solution?

eskil

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 = ?

rock.freak667

a²+a²=2a²
expand the right side and then simplify.

eskil

i don't believe that (a + 1)(a + 1) is 2a²
shouldn't that give a² + 2a +1 ??

NoMoreExams

[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.

snipez90

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.

eskil

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

NoMoreExams

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 | |.

eskil

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.