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Do you need to use the quadratic formula to solve for the unkown?

  1. May 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Lets say you have Ax + Bx^2 + C = 0

    Usually I would use the Quadratic formula in this situation.

    But what if you do this
    x( A + Bx ) + C = 0
    x( A + Bx ) = -C

    Couldn't you then say
    x = -C
    and
    A + Bx = -C
    x = (-C - A) / B

    ?
     
  2. jcsd
  3. May 27, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Well you could say that, but it won't be correct. If you sub x=-C you will get

    -C(A-BC) = -AC+BC2

    and that is not the right side of the equation in red.
     
  4. May 27, 2009 #3
    No. Factoring takes advantage of the property that if ab=0, then either a=0 or b =0. Furthermore, your answers don't solve the quadratic. Just plug them in.
     
  5. May 27, 2009 #4

    Mark44

    Staff: Mentor

    And to extend what Random Variable said, if you had ab = 1, then all you know about a and b is that they are reciprocals of one another. They could be 1 and 1, -1 and -1, 2 and 1/2, [itex]\sqrt{2}[/itex] and [itex]1/\sqrt{2}[/itex], or any other of an infinite number of pairs of numbers that multiply to 1.
     
  6. May 27, 2009 #5
    Thanks, this is all coming back now. :P

    So then my equation would work only if C = 0.


    @Mark44

    So if my C = -1 then saying
    Ax^2 + Bx = 1
    would be a valid way to solve the equation?

    Or are you saying
    Ax^2 + Bx + C = 1
    would be valid to solve for?
     
  7. May 27, 2009 #6

    Pengwuino

    User Avatar
    Gold Member

    No. Why do you think that? All that is telling you is that x and (A+Bx) are always -C when multiplied together if x is a solution to the quadratic formula. That doesn't tell you anything about what x actually is. Remember, the quadratic formula has 2 solutions so you have to get 2 numbers out of it for x.
     
  8. May 27, 2009 #7
    I believe that would be correct. If you had
    x (A+Bx) = 0

    then either (x) or (A+Bx) must be zero, or both. The two solutions would be
    x =0 and
    x=-A/B
     
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