Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$ 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
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!
$$8 = x\left [ 2-2\sqrt{5} \right ] + \left ( 1 - x \right )\left [ 2 + 2\sqrt{5} \right ]$$

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