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Math Help - Why does one have to divide through by "a" when completing the square?

  1. #1
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    Why does one have to divide through by "a" when completing the square?

    Essentially I can't figure out why you should divide out the "a" term in completing the square when you have the non-monic case?

    a(x-h)^2+k after all, is just 1(x-h)^2+k and if you don't have to divide when a=1, then why should the case change when a=n>1?
    Last edited by bournouli; March 30th 2011 at 12:46 PM.
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  2. #2
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    Have another go at posting this.
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  3. #3
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    We want to write ax^2+bx as a perfect square. The usual procedure says to add (and then subtract) (b/2)^2. But ax^2+bx+(b/2)^2 is only a perfect square when a=1.
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  4. #4
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    You don't have to divide by a, it just makes the calculations easier.
    (x+ b)^2= x^2+ 2bx+ b^2
    so if you have, say x^2+ 3x, you only have to think "2b= 3 so b= 3/2 and then b^2= 9/4- I have to add 9/4 to make this a perfect square".

    But (ax+ b)^2= a^2x^2+ 2abx+ b^2 so if you have, say, 2x^2+ 3x, you would have to think " a^2= 2 so a= \sqrt{2}, then 2ab= 2\sqrt{2}b= 3 so b= \frac{3}{2\sqrt{2}} and b^2= \frac{9}{8} so we need to add \frac{9}{8} to make that a perfect square".

    It is just simpler to factor out the "a" first.
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  5. #5
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    Dr. Steve,

    Why wont a quadratic be a perfect square when a doesn't equal 1? I can hypothesize that perhaps, from a geometric POV a square plus a square is more of a rectangle than anything else, and as the ax^2 term is a geometric square 2x^2 must be a rectangle so you have to "square" it, but I'm not sure if I'm right.
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    deleted - accidental double post
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  7. #7
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    The discriminant of ax^2+bx+(b/2)^2 is b^2-ab^2=b^2(1-a). Thus the discriminant is 0 if and only if b=0 or a=1. Thus, if b\ne 0 and a\ne 1, the quadratic equation ax^2+bx+(b/2)^2=0 has 2 distinct roots. Thus, when you factor ax^2+bx+(b/2)^2 you get two distinct factors. In order for the expression to be a perfect square, you need the same factor to appear twice.
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