# 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

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$(1-x)(2(1-\sqrt{5})) = 2+2\sqrt{5}-2x-2\sqrt{5}x$ right? – user13838 Sep 5 '11 at 19:56
Try gathering the terms containing $x$ on one side, and everything else on the other side. You should end up with an equation of the form $ax = b$ (no matter how "ugly" $a$ and $b$ look), whose solution is $\frac{b}{a}$. – Joel Cohen Sep 5 '11 at 20:02
@percusse: you have a sign wrong on lhs which should be $(1-x)(2(1+\sqrt{5})) = 2+2\sqrt{5}-2x-2x\sqrt{5}$ – Mark Bennet Sep 5 '11 at 20:04
There is a simple way: Just use Wolfram Alpha – FUZxxl Sep 5 '11 at 20:13
Wolfram alpha is great for verifieng but i need to know how to do this on my own in tests there is no wolfram – Jason Sep 5 '11 at 20:50

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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Hey thanks for the answer. Could you describe the "split out the terms" step I did not understand it – Jason Sep 5 '11 at 21:42
In the step which says "split out the terms" you see $$\frac{3-\sqrt{5}}{-2\sqrt{5}}$$ right?? the same can be expressed as the final step as two seperate fractions as $$-2\sqrt{5}$$ is common for both.Please note that after splitting the terms $$\sqrt{5}$$ in the numerater and $$\sqrt{5}$$ in the denominator get's canceled leaving us with +1/2 – alok Sep 5 '11 at 22:20
@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. – Steven Stadnicki Sep 5 '11 at 22:32
Ok Sweet thanks :) – Jason Sep 6 '11 at 20:30

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

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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This is incorrect. The original equation is linear and has only a single solution. Your extra solution candidate is caused by the unnecessary squaring (that's implication, not equivalence). – user6701 Sep 5 '11 at 20:04
@Tim : Agreed. The squaring of the right hand side seems dubious too (it should be $36 = 80 (1-x)^2$). – Joel Cohen Sep 5 '11 at 20:12
There was a mistake in step 3 which i corrected.I think now squaring both sides after subtracting 2 seems to be a legitimate step. – alok Sep 5 '11 at 20:17
@alok : Shouldn't there be a term in $x$ when you square the sum ? – Joel Cohen Sep 5 '11 at 20:27
@alok: Try to verify your solution by substituting $x$ for $\pm 1/\sqrt{5}$, and you'll see that at least one of those solutions is wrong. It is not correct that $x=y$ is equivalent to $x^2 = y^2$. – user6701 Sep 5 '11 at 20:37