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Quadratic Identities

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    2x^2 + 7x - 5 = A(x-1)^2 + Bx + C



    2. Relevant equations

    N/a

    3. The attempt at a solution

    This is an interesting equation as regardless of the whether I substitute x = 0 or x = 1 i still encounter the problem of not be able to make the calculation any easier. I have not been able to find any examples for guidence. Can someone assist.

    Cheers
     
    Last edited: Oct 11, 2009
  2. jcsd
  3. Oct 11, 2009 #2

    Dick

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    That's just an equation. What's the actual problem? Are you supposed to find A, B, and C given the equation is true for all x? That's just a guess.
     
  4. Oct 11, 2009 #3
    yes, sorry - need to find A, B and C

    Cheers


     
  5. Oct 11, 2009 #4

    Dick

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    Ok, then if ax^2+bx+c=0 for all x then a=0 and b=0 and c=0. Expand everything and move everything to one side. Set the coefficients of powers of x to zero and solve for A, B and C.
     
  6. Oct 12, 2009 #5

    HallsofIvy

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    Or do exactly what you suggested in your original post.

    If x= 0, this becomes -5= A+ C. If x= 1, this becomes 4= B+ C. Since there are three unknown numbers, you need a third equation: If x= -1, this becomes -10= 4A- B+ C. Three equations to solve for A, B, and C. From equation one, A= -5-C. From equation 2, B= 4- C. Replace A and B with those in the third equation and solve for C.
     
  7. Oct 12, 2009 #6
    I ended up with A=2, B=11 and C= -7

    Substituting the figures in to the equation - it balances with the R.H.S.

    I think that I have solved it - but happy for comment if I have it wrong.

    Cheers
     
  8. Oct 12, 2009 #7

    Borek

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    A(x-1)2 + Bx + C = Ax2 + (B-2A)x + A+C

    So at least A=2 is obviously OK, then A+C=2+C=-5, so C=-7. B-2A=B-4=7, so B is OK too.

    --
    methods
     
  9. Oct 12, 2009 #8
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