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Completing the Square

  1. May 30, 2009 #1
    Would completing the square be necessary to learn if I already know the quadratic formula? It seems to be a wasted amount of effort to learn completing the square when I already know the formula, or are there things that I need to know the squaring method for that the formula won’t work for?
     
  2. jcsd
  3. May 30, 2009 #2

    cepheid

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    You could have learned completing the square in the time it took you to type that message. Nothing in mathematics is a wasted effort. Your perspective seems to be a bit strange. The process of completing the square would not exist independently of other algebraic techniques if it was redundant.
     
  4. May 30, 2009 #3
    In general its more convenient to write things like

    [tex] x^{2} + 8x + 22 [/tex]

    as

    [tex] (x+4)^{2} +6 [/tex]

    which is done by completing the square, rather than finding out the values of x.

    The quadratic formula would give you [tex] \frac{-8 \pm \sqrt{8^{2} - 4(1)(22)}}{2(1)} [/tex] which gives non-real answers.

    With certain integrals that have fractions with polynomials in them, completing the square might be easier (especially if the values for x are decimals rather than whole numbers or as i n the example above).
     
  5. May 30, 2009 #4
    Also note that in protonchain's example, the geometric properties of the graph become immediately obvious in the completed version: the minimum/vertex of the function trivially occurs at x = -4 with value 6. The graph is simply the graph of y = x2 shifted to the left 4 units and up 6 units. The basic analysis of other quadratic forms benefit from factoring this way as well (conics). There are no square root extractions to worry about and no need to worry about complex numbers, just straightforward elementary arithmetic.
     
  6. May 30, 2009 #5
    Ok thanks for the help.
     
  7. May 30, 2009 #6
    The idea behind completing the square can be used in more general circumstances. For example, you can "complete the square" with matrices and vectors such as x*Mx + b*x where M is symmetric positive definite. This comes up in places like gaussian correlation, where M is the correlation matrix, and the expression is in an exponent to be integrated.
     
  8. May 31, 2009 #7
    An important fact here is to seperate out the roots. Putting the equation in the form X^2+BX+C = (X-a)(X-b) = X^2-(a+b)X +ab.

    Thus, it turns out that B^2-4C = (a-b)^2. And thus [tex]-B+\sqrt {(a-b)^2 }[/tex] allows us to seperate out the roots. This method seems to ignores the question of completing the square.
     
  9. May 31, 2009 #8
    The quadratic formula comes from completing the square, so in essence it's the same thing. For basic problems where you have to solve for some unknown it is a matter of preference, although as others have pointed out, one might be more benefitial in other circumstances.
     
  10. May 31, 2009 #9
    By the way, there are two quadratic formulas. One is used by mathematicians and is most commonly found in textbooks as the quadratic formula. But there's another equivalent one that programmers use because it behaves better when doing numerical calculations in floating point. I only mention this to argue that you shouldn't be content to stop learning, even when you think you know something.
     
  11. May 31, 2009 #10

    symbolipoint

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    Do you realize that Completing the Square has a geometric interpretation? This really IS "Completing The Square".
     
  12. May 31, 2009 #11
  13. Jun 4, 2009 #12
    Inverse Laplace transforms come to mind/
     
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