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I came up with this equation during my homework : $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$

My algebra is weak and I can't seem to find a way to solve for x nicely

Could someone please show me a decent way of doing this?

Thanks alot, Jason

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    @g.josh It appears you are trying to leave$a$comment, but I am not sure to whom you are addressing your comment, so I am leaving it as a comment on the main question.2012-10-17

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Generally the best way is to just plough through the algebra (and algebra gets quite a bit more advanced than this!) :

  • $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
  • $4=x(1-\sqrt{5})+(1-x)(1+\sqrt{5})$ (dividing through by $2$ simplifies a lot of the subsequent terms)
  • $4=x-x\sqrt{5}+1-x+\sqrt{5}-x\sqrt{5}$ (multiply out all the terms)
  • $4=x-x\sqrt{5}-x-x\sqrt{5}+1+\sqrt{5}$ (rearrange to get all the $x$ terms out front)
  • $4=x(1-\sqrt{5}-1-\sqrt{5})+1+\sqrt{5}$ (collect the x terms)
  • $4=x(-2\sqrt{5})+1+\sqrt{5}$ (simplify)
  • $(4-(1+\sqrt{5}))=x(-2\sqrt{5})$ (move the constant term to the left)
  • $3-\sqrt{5} = x(-2\sqrt{5})$ (simplify the left)
  • $x=(3-\sqrt{5})/(-2\sqrt{5})$ (divide both sides by $-2\sqrt{5}$)
  • $x=3/(-2\sqrt{5}) + {1\over2}$ (split out the terms)
  • $\displaystyle{x=-{3\sqrt{5}\over 10} + {1\over2}}$ (multiply the numerator and denominator of the first part through by $\sqrt{5}$)

Of course, I strongly recommend plugging this $x$ in to confirm that it satisfies your initial equation!

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    @Jason alok has it right - there's a small missing intermediate step here, which is to write $(3-\sqrt{5})/(-2\sqrt{5})$ as $(3/(-2\sqrt{5})) - (\sqrt{5}/(-2\sqrt{5}))$, and then simplify the latter by dividing through by $-\sqrt{5}$. You'll definitely get used to skipping through these sorts of intermediate steps with time.2011-09-05
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$8 = x\left [ 2-2\sqrt{5} \right ] + \left ( 1 - x \right )\left [ 2 + 2\sqrt{5} \right ]$

use foil

$8 = 2x - 2x\sqrt{5} + 2 + 2\sqrt{5} - 2x - 2x\sqrt{5}$

$8 = -4x\sqrt{5} + 2\sqrt{5} + 2$

subtract 2 and square both sides

$36 = 16x^25 + 20$

subtract 20 from both sides to obtain

$80x^2 = 16$

Now divide both sides by 80 and you get $x^2 = 16/80$

therefore $x = \pm 1 / \sqrt{5}$

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    Your squaring both sides is wrong, too. $(2\sqrt{5}-4x\sqrt{5})^2=5(4-16x+16x^2)$. Now you should just move the terms without $x$ to the right and the term with $x$ to the left and divide.2011-09-05