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Thread: 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; Mar 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 $\displaystyle ax^2+bx$ as a perfect square. The usual procedure says to add (and then subtract) $\displaystyle (b/2)^2$. But $\displaystyle ax^2+bx+(b/2)^2 $is only a perfect square when $\displaystyle a=1$.
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  4. #4
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    You don't have to divide by a, it just makes the calculations easier.
    $\displaystyle (x+ b)^2= x^2+ 2bx+ b^2$
    so if you have, say $\displaystyle x^2+ 3x$, you only have to think "2b= 3 so b= 3/2 and then $\displaystyle b^2= 9/4$- I have to add 9/4 to make this a perfect square".

    But $\displaystyle (ax+ b)^2= a^2x^2+ 2abx+ b^2$ so if you have, say, $\displaystyle 2x^2+ 3x$, you would have to think "$\displaystyle a^2= 2$ so $\displaystyle a= \sqrt{2}$, then $\displaystyle 2ab= 2\sqrt{2}b= 3$ so $\displaystyle b= \frac{3}{2\sqrt{2}}$ and $\displaystyle b^2= \frac{9}{8}$ so we need to add $\displaystyle \frac{9}{8}$ 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 $\displaystyle ax^2+bx+(b/2)^2$ is $\displaystyle b^2-ab^2=b^2(1-a)$. Thus the discriminant is 0 if and only if b=0 or a=1. Thus, if $\displaystyle b\ne 0$ and $\displaystyle a\ne 1$, the quadratic equation $\displaystyle ax^2+bx+(b/2)^2=0$ has 2 distinct roots. Thus, when you factor $\displaystyle ax^2+bx+(b/2)^2$ 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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