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Quadratic Equations

  1. Jul 15, 2010 #1
    Need help solving some. I'll put up one for now.

    a/b-x + b/a-x = 2

    I'm just looking for how to solve it, because I always reach

    2x^2 - 3ax - 3bx + a^2 + b^2 + 2ab = 0

    and get stuck. Please show me your steps.

    Thanks!
     
  2. jcsd
  3. Jul 15, 2010 #2
    That = 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0

    Do you know the quadratic equation?
     
  4. Jul 15, 2010 #3
    I know quadratic equations, but I'm getting stuck here, because I do not know how to solve for x.

    I did establish the fact that 2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0
    is correct because it follows ax^2 + bx + c.

    Could you please help me out?
     
  5. Jul 15, 2010 #4

    Mark44

    Staff: Mentor

    What this means is probably different from what you intended. When you write fractions in a single line of text, use parentheses if the numerator or denominator have more than one term.

    What you wrote would be interpreted as this:
    [tex]\frac{a}{b - x} + \frac{b}{a - x} = 2[/tex]
    I get something different - different coefficients for the two terms in x, and different signs for the a^2 and b^2 terms. Check your work.
     
  6. Jul 15, 2010 #5

    Mark44

    Staff: Mentor

    2x^2 -3(a+b)x + (a^2 + 2ab + b^2) = 0 IS a quadratic equation. The question should be: Do you know the quadratic formula?

    The quadratic formula gives solutions to the equation ax^2 + bx + c = 0. Since both the quadratic formula and the equation above use a and b, be careful in what you call a, b, and c in the quadratic formula.

    As already noted, some of the coefficients in the OP's equation are incorrect.

    Also, since the equation in the original post involved division by a - x and b - x, it should be stated explicitly that x can't be a, and x can't be b.
     
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