Completing the Square

TL;DR

Completing the square is a super useful technique to rewrite quadratic expressions into a special form that makes solving them or graphing them much easier. It turns a standard quadratic expression into a perfect square trinomial plus a constant, which helps us find the vertex of a parabola or solve equations that aren't easily factorable. This skill is crucial for understanding parabolas and preparing for more advanced math topics.

1. The Mental Model

Imagine you have a jigsaw puzzle piece that's almost a perfect square. Completing the square is like adding just the right small piece to make it a perfect square, then adjusting for what you added. You're transforming an expression into a more useful shape.

2. The Core Material

Completing the square is a method used to convert a quadratic expression of the form $ax^2 + bx + c$ into the form $a(x-h)^2 + k$. This new form, called vertex form, directly tells you the vertex of the parabola $(h, k)$, which is incredibly helpful for graphing and finding minimum/maximum values.

Let's focus on the case where $a=1$ first, as it's the most common starting point.

The Basic Idea: Creating a Perfect Square Trinomial

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A perfect square trinomial is a trinomial that results from squaring a binomial. For example:
* $(x+5)^2 = x^2 + 10x + 25$
* $(x-3)^2 = x^2 - 6x + 9$

Notice a pattern? In $x^2 + bx + c$, if it's a perfect square, $c$ is always $(\frac{b}{2})^2$.
So, if you have $x^2 + bx$, you need to add $(\frac{b}{2})^2$ to make it a perfect square trinomial.

Steps for Completing the Square (when $a=1$)

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1. Isolate the $x^2$ and $x$ terms: If you have an equation like $x^2 + bx + c = 0$, move the constant term $c$ to the other side: $x^2 + bx = -c$. If it's just an expression, group $x^2 + bx$.
2. Find the magic number: Take half of the coefficient of your $x$ term ($b$), and then square it. This is $(\frac{b}{2})^2$.
3. Add the magic number: Add this number to both sides of the equation (if solving an equation) or add and subtract it (if transforming an expression) to keep things balanced.
4. Factor the perfect square: The $x^2 + bx + (\frac{b}{2})^2$ part will now factor into $(x + \frac{b}{2})^2$.
5. Simplify: Combine the constant terms.

Let's visualize this process:

graph TD
    A["Start with x^2 + bx + c"] --> B{"Is 'a' equal to 1?"}
    B -- Yes --> C["Group (x^2 + bx) + c"]
    C --> D["Calculate (b/2)^2"]
    D --> E["Add and subtract (b/2)^2 inside the parentheses: (x^2 + bx + (b/2)^2) + c - (b/2)^2"]
    E --> F["Factor the perfect square: (x + b/2)^2 + c - (b/2)^2"]
    B -- No --> G["Factor 'a' out of the x^2 and x terms: a(x^2 + (b/a)x) + c"]
    G --> H["Inside parentheses, calculate ((b/a)/2)^2 = (b/2a)^2"]
    H --> I["Add and subtract (b/2a)^2 *inside* the parentheses: a(x^2 + (b/a)x + (b/2a)^2 - (b/2a)^2) + c"]
    I --> J["Factor the perfect square and distribute 'a' to the subtracted term: a(x + b/2a)^2 - a(b/2a)^2 + c"]
    J --> K["Simplify constants"]
    F --> K
    K --> L["Result: a(x-h)^2 + k or solve for x"]

When 'a' is Not 1

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If your quadratic expression is $ax^2 + bx + c$ where $a \neq 1$, you need an extra step:

1. Factor out 'a': Factor $a$ out of only the $x^2$ and $x$ terms: $a(x^2 + \frac{b}{a}x) + c$.
2. Complete the square inside the parentheses: Now, use the steps from before on the expression inside the parentheses, using $\frac{b}{a}$ as your new $b$:
   • Find $(\frac{b/a}{2})^2 = (\frac{b}{2a})^2$.
   • Add and subtract this term inside the parentheses.
     $a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2) + c$
3. Move the subtracted term out: When you move the subtracted term from inside the parentheses to outside, remember to multiply it by $a$.
   $a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2) - a(\frac{b}{2a})^2 + c$
4. Factor and simplify:
   $a(x + \frac{b}{2a})^2 - a\frac{b^2}{4a^2} + c$
   $a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$
   $a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}$

This final form is $a(x-h)^2 + k$, where $h = -\frac{b}{2a}$ and $k = \frac{4ac - b^2}{4a}$.

3. Worked Example

Let's transform the quadratic expression $2x^2 + 12x + 13$ into vertex form by completing the square.

1. Factor out 'a' from the $x^2$ and $x$ terms:
   $2(x^2 + 6x) + 13$

2. Find the magic number for the part inside the parentheses:
   The new $b$ is $6$.
   $(\frac{6}{2})^2 = 3^2 = 9$.

3. Add and subtract this number inside the parentheses:
   $2(x^2 + 6x + 9 - 9) + 13$

4. Move the subtracted term outside the parentheses, remembering to multiply by 'a':
   $2(x^2 + 6x + 9) - 2(9) + 13$
   $2(x^2 + 6x + 9) - 18 + 13$

5. Factor the perfect square trinomial and simplify the constants:
   $2(x + 3)^2 - 5$

So, $2x^2 + 12x + 13$ is equivalent to $2(x+3)^2 - 5$. From this vertex form, we can immediately tell the vertex of the parabola is $(-3, -5)$.

4. Key Takeaways

• Completing the square changes $ax^2 + bx + c$ into vertex form $a(x-h)^2 + k$.
• The key is to create a perfect square trinomial, $(x \pm \text{something})^2$.
• For $x^2 + bx$, the missing piece to make it a perfect square is always $(\frac{b}{2})^2$.
• If $a \neq 1$, factor $a$ out of the $x^2$ and $x$ terms first, then complete the square inside.
• Remember to multiply the subtracted term by $a$ when moving it outside the parentheses.
• The vertex form $a(x-h)^2+k$ directly shows the parabola's vertex at $(h,k)$.
• This method is useful for solving quadratic equations that don't easily factor.

Common mistakes to avoid:
* Forgetting to balance the equation by adding the same number to both sides (when solving) or adding and subtracting (when transforming an expression).
* When $a \neq 1$, forgetting to multiply the subtracted constant by $a$ when it's moved outside the parentheses.
* Incorrectly calculating $(\frac{b}{2})^2$, especially with negative $b$ values (remember, squaring always results in a positive number).
* Not factoring $a$ out completely from both $x^2$ and $x$ terms when $a \neq 1$.

5. Now Try It

Transform the quadratic expression $3x^2 - 18x + 20$ into vertex form using the completing the square method.

What to do: Follow the steps shown in the Core Material and the Worked Example. Focus on factoring out the 'a' term first, then completing the square inside the parentheses, and finally simplifying.

What success looks like: Your final answer should be in the form $a(x-h)^2 + k$, and you should be able to identify the vertex $(h, k)$ from your result. Take about 15 minutes to work through it.