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Completing the square

Warmup: Solve: $x^{2} + 6 x + 2 = 0$ .

Hard or impossible to factor.

Look at $x^{2} + 6 x$ . (Ignore the 2 for the moment.)

Notice this:

$\cdot$	$x$	$+$	3
$x$	$x^{2}$		$3 x$
$+$
3	$3 x$		$3^{2}$

This gives us

\begin{array}{rcl} {(x + 3)}^{2} & = & x^{2} + 3 x + 3 x + 3^{2} \\ = & x^{2} + 6 x + 3^{2} \end{array}

Looking at it another way, ${(x + 3)}^{2}$ is almost equal to $x^{2} + 6 x$ except for that pesky $3^{2}$ .

2nd trick: To solve $x^{2} + 6 x + 2 = 0$ , (i.e. to find the zeros of the function $x^{2} + 6 x + 2$ :

\begin{array}{rcl} x^{2} + 6 x + 2 & = & 0 \\ x^{2} + 6 x + 3^{2} + 2 & = & 3^{2} \\ x^{2} + 6 x + 3^{2} & = & 3^{2} - 2 \\ {(x + 3)}^{2} & = & 3^{2} - 2 \\ x + 3 & = & \pm \sqrt{3^{2} - 2} \\ x & = & - 3 \pm \sqrt{3^{2} - 2} \\ = & - 3 \pm \sqrt{7} \end{array}

The thing on the right is messy, but it's just a number (no $x$ ).

Check the plus version (minus is similar):

\begin{array}{rcl} x^{2} + 6 x + 2 & = & {(- 3 + \sqrt{7})}^{2} + 6 (- 3 + \sqrt{7}) + 2 \\ = & 9 - 3 \sqrt{7} - 3 \sqrt{7} + 7 - 18 + 6 \sqrt{7} + 2 \\ = & 9 + 7 - 18 + 2 - 6 \sqrt{7} + 6 \sqrt{7} \\ = & 0 \end{array}

This can be messy but it always works!

If $a \neq 1$ , just divide both sides by $a$ first:

\begin{array}{rcl} 3 x^{2} - 6 x + 12 & = & 0 \\ x^{2} - 2 x + 4 & = & 0 \\ x^{2} - 2 x & = & - 4 \\ x^{2} - 2 x + {(- 1)}^{2} & = & - 4 + {(- 1)}^{2} \\ {(x - 1)}^{2} & = & - 3 \\ x - 1 & = & \pm \sqrt{- 3} \\ x & = & 1 \pm \sqrt{- 3} \\ x & = & 1 \pm i \sqrt{3} \end{array}

You can get a quadratic function into vertex form by completing the square with the function definition:

\begin{array}{rcl} y & = & x^{2} - 8 x + 11 \\ y + {(- \frac{8}{2})}^{2} & = & x^{2} - 8 x + {(- \frac{8}{2})}^{2} + 11 \\ y + 16 & = & x^{2} - 8 x + 16 + 11 \\ y + 16 & = & {(x - 4)}^{2} + 11 \\ y & = & {(x - 4)}^{2} - 5 \end{array}

The vertex is $(4, 5)$ .

The minimum of a quadratic whose $a$ term is positive, or the maximum of a quadratic whose $a$ term is negative, the vertex. So to find the max or min, just find the vertex.