simple solution?


by eskil
Tags: simple, solution
eskil
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#1
Aug20-08, 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:

a2 + a2 = (a + 1)2

a = ?
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rock.freak667
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#2
Aug20-08, 12:29 PM
HW Helper
P: 6,214
a2+a2=2a2
expand the right side and then simplify.
eskil
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#3
Aug20-08, 12:37 PM
P: 4
i don't believe that (a + 1)(a + 1) is 2a2
shouldn't that give a2 + 2a +1 ??

NoMoreExams
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#4
Aug20-08, 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.
snipez90
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#5
Aug20-08, 01:12 PM
P: 1,106
Quote Quote by rock.freak667 View Post
a2+a2=2a2
expand the right side and then simplify.
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
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#6
Aug20-08, 01:22 PM
P: 4
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
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#7
Aug20-08, 01:29 PM
P: 626
Quote Quote by eskil View Post
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
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
eskil is offline
#8
Aug20-08, 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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