## Completing the Square: Two Forms – Two Methods

Consider two expansions of $(a+b)^2$$a^2+2ab+b^2$ and $a^2+b(2a+b)$.  These expressions are algebraically equivalent yet each contains an implicit point of view that leads to a different method for completing the square.

Let’s look at a specific problem.  Express $y = x^2+6x+10$ in the standard form $y = a(x-h)^2+k$.

### Usual Method

Here $a$ is $1$ so we try to look at $x^2+6x+10$ as an expression like $(x-h)^2 = x^2-2xh+h^2$.  We already have $x^2$ so $6x = -2xh$ and $h=-3$ , $h^2=9$.  So $x^2+6x+10 =x^2+6x+9+10-9$ and we end up with $y=(x+3)^2+1$.  Note the usual trick of adding and subtracting $9$.  Rather unaesthetic and hard for students to remember.

### Another Method

Look at $x^2+6x+10$ as $x^2-h(2x-h)$.  We have the $x^2$ already so $6x+10=-h(2x-h)$$h$ must be $-3$ so $6x+10=-(-3)(2x-(-3)=6x+9$.  Oops.  We have $1$ leftover ($10-9$) so $y=(x+3)^2+1$.   As described this method is nearly as awkward as the usual method.   This is because the way  the explanation was written obscures the steps. The “Another Method” uses the steps for finding the square root of any polynomial.  This method can be found in A College Algebra by G.A. Wentworth [1].

The basic idea is that we are trying to find $\sqrt{N}$, where $N$ is some number or polynomial.  We know that $\sqrt{N} = a+b$ and somehow we know $a$ , the “larger” part.  We want to find $b$ or at least refine our knowledge about $b$$N = a^2+b(2a+b)$.  So we estimate $b$ with $\frac{N-a^2}{2a}$, calculate $N-(a^2+b(2a+b)$ and continue,  using $a+b$ as our new $a$ if the expression is not equal to zero.  This is the old-fashioned (and slow) way to find the square root of  a number by hand.  See http://www.homeschoolmath.net/teaching/square-root-algorithm.php for an example.

Here is the method applied to our example polynomial using the usual long division-type layout.This is a very clean procedure for completing the square.  The “remainder” appears naturally without the trick of adding and subtracting $h^2$.  It has the additional advantage of generalizing to finding the square root of any polynomial or number.

The algorithm is fun to use.  Try $\sqrt{x^2+7x+10}$ and $\sqrt{3x^2+5x+10}$ with first step $a = \sqrt{3}x$ and compare to $\sqrt{3x^2+5x+10} = \sqrt{3}\sqrt{x^2+\frac{5}{3}x+\frac{10}{3}}$.

There is a comparable procedure for finding cube roots in Wentworth’s text.  I spent some time trying to solve quartic and cubic polynomial equations by taking square and cube roots but didn’t find any new cool  easy procedures.

Now to make a meta-point.  Different forms on an expression can give one different insights into the structure of a problem.  It is sometimes useful to look at a different version of an expression and construct a narrative that “explains” the form.

[1]  G.A. Wentworth, A College Algebra,Revised Edition, Ginn and Company, 1902.