1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 27, 2009 #1

    yougene

    User Avatar

    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

    ?
     
    yougene, May 27, 2009
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  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.
     
    rock.freak667, May 27, 2009
  4. May 27, 2009 #3

    Random Variable

    User Avatar

    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.
     
    Random Variable, May 27, 2009
  5. May 27, 2009 #4

    Mark44

    User Avatar
    Insights Author

    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.
     
    Mark44, May 27, 2009
  6. May 27, 2009 #5

    yougene

    User Avatar

    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?
     
    yougene, May 27, 2009
  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.
     
    Pengwuino, May 27, 2009
  8. May 27, 2009 #7

    Chrisas

    User Avatar

    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
     
    Chrisas, May 27, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook