- #1

- 4

- 0

a

^{2}+ a

^{2}= (a + 1)

^{2}

a = ?

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- Thread starter eskil
- Start date

- #1

- 4

- 0

a

a = ?

- #2

rock.freak667

Homework Helper

- 6,223

- 31

a^{2}+a^{2}=2a^{2}

expand the right side and then simplify.

expand the right side and then simplify.

- #3

- 4

- 0

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

shouldn't that give a^{2} + 2a +1 ??

shouldn't that give a

- #4

- 623

- 0

[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:

- #5

- 1,101

- 3

a^{2}+a^{2}=2a^{2}

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.

- #6

- 4

- 0

a

simplified it to a quadraticequation

0 = -a

a

a

a

which gives a = 2,41

- #7

- 623

- 0

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 | |.

- #8

- 4

- 0

I still think that using the quadratic equation is the simplest way of solving it.

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