Completing the Square

Another method of solving a quadratic is to complete the square. This changes the form of the quadratic from

(1)   \begin{equation*} y = ax^2 + bx + c \end{equation*}

Into the form

(2)   \begin{equation*} y = a(x - h)^2 + k \end{equation*}

We can use the square of a binomial:

(3)   \begin{equation*} (x + a)^2 = x^2 + 2ax + a^2 \end{equation*}

Using this we can see that for any equation of the form y=x^2 + bx + c, in other words, where a=1 we can form a square:

(4)   \begin{equation*} \left( x + \frac{1}{2} b \right) ^2 = x^2 + bx + \frac{1}{4} b^2 \end{equation*}

So:

(5)   \begin{equation*} x^2 + bx + c = \left( x + \frac{1}{2} b \right)^2 + k \end{equation*}

Where k is:

(6)   \begin{equation*} k = c - \frac{b^2}{4} \end{equation*}

So this gives us:

(7)   \begin{equation*} x^2 + bx + c = \left( x + \frac{1}{2} b \right)^2 + c - \frac{b^2}{4} \end{equation*}

We can expand this to form a general case where

(8)   \begin{equation*} ax^2 + bx + c \end{equation*}

into the form

(9)   \begin{equation*} a(x - h)^2 + k \end{equation*}

So the general case is:

(10)   \begin{equation*} ax^2 + bx + c = a(x - h)^2 + k \end{equation*}

Where

(11)   \begin{equation*} h = -\frac{b}{2a} \end{equation*}

and

    \[ \begin{split} k & = c - ah^2 \\ & = c - a\left\frac{b}{2a}\right^2 \\ & = c - a\left(\frac{b^2}{4a^2}\right) \\ & = c - \frac{b^2}{4a} \end{split} \]

The completed square form is useful as the values of h and k give us the Cartesian coordinates of the stationary point of the parabola, so we have a local minimum or maximum at (h,k).

So for example, take the quadratic:

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So completing the square with a=6, b=5 and c=-6, we have:

    \[ \begin{split} h & = -\frac{b}{2a} \\ & = -\frac{5}{2 \cdot 6} \\ & = -\frac{5}{12} \end{split} \]

and

    \[ \begin{split} k & = c - \frac{b^2}{4a} \\ & = -6 - \frac{5^2}{4 \cdot 6} \\ & = -6 - \frac{25}{24} \\ & = -\frac{169}{24} \\ \end{split} \]

So the local minimum in this case is at \left(-\frac{5}{12},-\frac{169}{24}\right), and the completed square form is:

(12)   \begin{equation*} 6x^{2} + 5x - 6 = 6\left(x + \frac{5}{12}\right)^2 - \frac{169}{24} \end{equation*}

 

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