2
$\begingroup$

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

  • 0
    $(1-x)(2(1-\sqrt{5})) = 2+2\sqrt{5}-2x-2\sqrt{5}x$ right?2011-09-05
  • 3
    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 $\frac{b}{a}$.2011-09-05
  • 0
    @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}$2011-09-05
  • 0
    There is a simple way: Just use [Wolfram Alpha](http://wolframalpha.com)2011-09-05
  • 2
    Wolfram alpha is great for verifieng but i need to know how to do this on my own in tests there is no wolfram2011-09-05
  • 0
    @Mark : you are right, I wasn't careful enough, sorry about that.2011-09-05
  • 0
    hai this is josh from jntu(javaharlalnehru technological science of hyderabad).and i want to more impermation about your formula's.please send me all information about this.2012-10-17
  • 0
    @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

2 Answers 2

6

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!

  • 0
    Hey thanks for the answer. Could you describe the "split out the terms" step I did not understand it2011-09-05
  • 1
    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 +1/22011-09-05
  • 2
    @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
-3

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

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

  • 4
    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).2011-09-05
  • 0
    @Tim : Agreed. The squaring of the right hand side seems dubious too (it should be $36 = 80 (1-x)^2$).2011-09-05
  • 0
    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.2011-09-05
  • 1
    @alok : Shouldn't there be a term in $x$ when you square the sum ?2011-09-05
  • 0
    @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 $x^2 = y^2$.2011-09-05
  • 0
    Hmm..That's a good point.Seems to me even though this equation is quite elementary ,it's pretty involved.It's little bit hard.2011-09-05
  • 0
    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