
#1
Aug2008, 12:13 PM

P: 4

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 = ? 



#2
Aug2008, 12:29 PM

HW Helper
P: 6,213

a^{2}+a^{2}=2a^{2}
expand the right side and then simplify. 



#3
Aug2008, 12:37 PM

P: 4

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



#4
Aug2008, 12:41 PM

P: 626

simple solution?
[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.




#5
Aug2008, 01:12 PM

P: 1,106





#6
Aug2008, 01:22 PM

P: 4

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 



#7
Aug2008, 01:29 PM

P: 626

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  . 



#8
Aug2008, 03:51 PM

P: 4

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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