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Solving a quadratic

  1. Aug 25, 2004 #1
    in a maths book i have there is an example of solving a quadratic. part of the process reads:

    y^2 - 12y + 32 = 0


    implies that


    (y-8) (y-4) = 0



    i don't understand how this second equation was reached based on the first one.

    could someone add in the steps that gets me from the first equation to the implication.

    thanks.
     
  2. jcsd
  3. Aug 25, 2004 #2

    jcsd

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    expand:

    (y + a)(y + b)

    and you get:

    y^2 + (a+b)y + ab

    so from your equation above you know that:

    a + b = -12

    and

    ab = 32

    as

    -8 + -4 = -12

    and

    -8*-4 = 32

    a and b must be -8 and -4.
     
  4. Aug 25, 2004 #3

    xt

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    i don't think you can, it is obvious.

    if you actually want the logics of thinking its like this:

    we want to express the quadratic as (y-a)(y-b),

    now a*b = 32

    and a + b = 12

    so try all the integers you can think of, you'll find 8 and 4 does.
     
  5. Aug 25, 2004 #4

    arildno

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    A simple way uses the idea of "completing the square:
    [tex]y^{2}-12y+32=0[/tex]
    Now, regarding -12=2*(-6), we add 0 to our equation in this manner:
    [tex]y^{2}-12y+32+(-6)^{2}-(-6)^{2}=0[/tex]
    Or, equivalently, for the Left Hand Side:
    [tex]y^{2}-12y+32+(-6)^{2}-(-6)^{2}=(y^{2}-2*6y+6^{2})+(32-6^{2})=(y-6)^{2}-4[/tex]
    Furthermore, since [tex]4=2^{2}[/tex] we have:
    [tex](y-6)^{2}-4=(y-6)^{2}-2^{2}=(y-6+2)(y-6-2)=(y-4)(y-8)[/tex]

    Finally, by setting this expression (which is equivalent to our original left hand side) equal to our originil right hand side (that is,0) we gain:
    [tex](y-4)(y-8)=0[/tex]
    as required.
     
  6. Aug 25, 2004 #5
    thanks people.
    it was the bit i have highlighted in bold that i had forgotten about when trying to work it out.

    y^2 + (a+b)y + ab

    thanks again.
     
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