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Two equations to solve simultaneously

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data
    2 = c(1+sqrt(2)) + d(1-sqrt(2))
    5 = c(1+sqrt(2))2 + d(1-sqrt(2))2


    3. The attempt at a solution

    Ive tried finding an expression for c, and replacing c in the other equation, but it gives me the wrong answer... what should i do?

    The answer should be c=(sqrt(2)+2)/4
     
    Last edited by a moderator: Oct 3, 2014
  2. jcsd
  3. Oct 3, 2014 #2

    pasmith

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    Homework Helper

    Show your working. We can't help you if we can't see where you might have gone wrong.
     
  4. Oct 3, 2014 #3

    HallsofIvy

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    You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with [tex]A= 1+ \sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex]. Yes, you could solve the first equation for c, [tex]c= (2- Bd)/A[/tex], and put that into the second equation: A^2[(2- Bd)/A)+ B^2d= A(2- Bd)+ B^2d= 2A- ABd+B^2d= 5 so (B^2- AB)d= 5- 2A and d= (5- 2A)/(B^2- AB)

    Personally, I would have multiplied the first equation by A to get A^2c+ ABd= 2A and then subtract that from the second equation: (B^2- AB)d= 5- 2A which immediately gives d= (5- 2A)/(B^2- AB). Similarly, multiply the first equation by B to get ABc+ B^2d= 2B and subtract that from the second equation to get (A^2- AB)c= 5- 2B so that c= (5- 2B)/(A^2- AB).

    Now put [tex]A= 1+\sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex].
     
  5. Oct 3, 2014 #4
    Thanks! I got the correct answer using d = (5- 2A)/(B^2- AB). Turns out I've been doing a mistake before i got to this part...
     
  6. Oct 3, 2014 #5
    Which, of course, should simplify to...

    c = (2 + Sqrt[2])/4

    d = (2 - Sqrt[2])/4
     
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