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Proving the Quadratic Equation

  1. Dec 18, 2005 #1
    Hi! I'm new here so if this has been covered please accept my apologies.
    I'm having trouble proving the equation from:
    ax²+bx+c=0 to the solutions x=(-b(+/-)√b²-4ac)/2a
    I've been told to complete the sqaure of ax²+bx+c, then rerrange to give x, but I just can't get it!
    If you could show the steps you took to get there it would be very helpful too.
    Any help would be appreciated. Thanks :smile:
     
  2. jcsd
  3. Dec 18, 2005 #2

    matt grime

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    what do you get after completing the square, assuming that you know how to complete the square? and homework ought to go in the homework section.
     
  4. Dec 18, 2005 #3
    I'm sorry. Its not homework. I just need to know how for an exam. When I complete the square I get:
    Well, this is what i've got on my sheet. I just don't understand step three!

    ax²+bx+c=0
    1 a[x²+bx+c]=0 --> x²+(b/a)x+c/a=0
    2 (x+b/2a)²+c/a-b²/4a²=0 --> (x+b/2a)²=b²/4a²-c/a
    3 (x+b/2a)=((+/-)√b²-4ac)/2a --> x=-b/2a((+/-)√b²-4ac)/2a
    4 x=(-b(+/-)√b²-4ac)/2a

    I don't understand where on the sheet it got from step 2 to step 3! That is from
    (From step 2): =b²/4a²-c/a to
    (From step 3): =((+/-)√b²-4ac)/2a


    Cheers
     
    Last edited: Dec 18, 2005
  5. Dec 18, 2005 #4

    uart

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    Just put the RHS over a common denominator of [tex]4a^2[/tex] and then take the square-root of both sides.
     
  6. Dec 18, 2005 #5
    RHS :confused: - Edit, right hand side!
     
  7. Dec 18, 2005 #6
    I understand it now! Thanks alot! It actually makes alot more sense now! Cheers!
     
  8. Dec 19, 2005 #7

    mathwonk

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    I never could follow that in high school. Now I am not sure why not, too many formulas, or fractions, i guess.

    lets try it with no "a", just x^2 + bx + c. then we have to separate out the first two terms from the last:

    x^2 + bx +c, then we fill in whatever makes it a square.

    thats it i hate fractions, so lets begin with x^2 + 2bx +c,

    then we get x^2 + 2bx +c = x^2 + 2bx + b^2 - b^2 + c

    = (x+b)^2 + c - b^2. so now if it all = 0, we get (x+b)^2 = b^2 -c,

    so x+b = + or - sqrt(b^2-c), and hence x = -b (+or-) sqrt(b^2-c).

    now isn't that easier?

    then you can solve the gbeneral case ax^2 + bx + c by changing it nito thius one:

    i.e. divide by a, getting x^2 + (b/a)X + c/a. then get a 2 in there in the middle as follows:


    x^2 + 2(b/2a)x + c/a. then this =0 if and only if

    x = -(b/2a) (+or-) sqrt(b^2/4a^2 - c/a)

    = -(b/2a) (+or-) sqrt(b^2/4a^2 - 4ac/4a^2)

    = -(b/2a) (+or-) sqrt(b^2 - 4ac)/sqrt(4a^2)

    = -(b/2a) (+or-) sqrt(b^2 - 4ac)/2a

    = [-b (+or-) sqrt(b^2 - 4ac)]/2a.


    what a messy formula, i like mine much better.
     
  9. Dec 19, 2005 #8

    mathwonk

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    e.g. to solve 4x^2 + 6x -1 = 0, change it to
    x^2 +(3/2)x -1/4 = x^2 + 2(3/4)x - 1/4, and the solution by my formula
    is x = -(3/4) (+or-) sqrt(9/16 + 1/4) = -(3/4) (+or-)sqrt(13)/2.
    is that right? it seems nicer than the usual one to me anyway.
    my advice to young learners is to have somewhat less respect for tradition.
     
    Last edited: Dec 19, 2005
  10. Dec 19, 2005 #9

    mathwonk

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    or is the usual one better? i get

    x =[-6 (+or-) sqrt(36+16)]/8 i guess thats the same, and fewer fractions too! ok tradition has something going for it.
     
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