MPM2D Completing the Square: Notes and Practice

A ball is thrown upward from a 2 m balcony, and its height in metres after $t$ seconds is $h = - 5 t^{2} + 20 t + 2$ . How high does it go? The maximum is the $y$ -coordinate of the vertex, but standard form hides the vertex. Factoring will not help either: no pair of whole numbers works for $- 5 t^{2} + 20 t + 2$ . What you need is a way to turn any standard form into vertex form. That method is called completing the square, and by the end of this lesson you will show that $h = - 5 (t - 2)^{2} + 22$ , so the ball reaches $22$ m after $2$ seconds.

The idea rests on perfect square trinomials, the expressions you get by squaring a binomial. Expanding $(x + 3)^{2}$ gives $x^{2} + 3 x + 3 x + 9 = x^{2} + 6 x + 9$ . Look at how the numbers are linked: the middle coefficient $6$ is double $3$ , and the constant $9$ is $3^{2}$ . So if you see $x^{2} + 6 x$ , you can tell it needs $+ 9$ to become a perfect square: take half the coefficient of $x$ , then square it. In general:

(x + d)^{2} = x^{2} + 2 d x + d^{2} ⟹ x^{2} + b x + (\frac{b}{2})^{2} = (x + \frac{b}{2})^{2}

Fig 1Algebra tiles for x² + 6x + 4. Left: one x-squared tile, six x-tiles and four unit tiles. Right: the x-tiles split into two groups of three, one along the right side and one along the bottom of the x-squared tile; the corner needs 9 unit tiles but only 4 are available, so x² + 6x + 4 = (x + 3)² − 5.

Algebra tiles make this visible. The big square tile has area $x \times x = x^{2}$ , each long tile has area $x \times 1 = x$ , and each small tile has area $1$ . To arrange $x^{2} + 6 x + 4$ into a square, split the six $x$ -tiles in half: three along the right side of the big square and three along the bottom. That builds a square $x + 3$ units on each side, except for a $3 \times 3$ corner, which needs $9$ unit tiles. You only have $4$ , so you are $5$ short. The tiles you have make "a full $(x + 3)$ by $(x + 3)$ square, minus 5": $x^{2} + 6 x + 4 = (x + 3)^{2} - 5$ .

Completing the square, step by step

y = x^{2} + 6 x + 4 = x^{2} + 6 x + 9 - 9 + 4 = (x^{2} + 6 x + 9) - 5 = (x + 3)^{2} - 5 half of 6 is 3, and 3^{2} = 9 group the perfect square vertex (- 3, - 5)

Adding $9$ and subtracting $9$ adds zero in total, so the relation never changes; it is only rewritten. When the leading coefficient $a$ is not 1, there is one extra step at the start and one extra multiplication near the end. Here is $y = 3 x^{2} - 12 x + 1$ worked in full.

y = 3 x^{2} - 12 x + 1 = 3 (x^{2} - 4 x) + 1 = 3 (x^{2} - 4 x + 4 - 4) + 1 = 3 (x^{2} - 4 x + 4) - 12 + 1 = 3 (x - 2)^{2} - 11 factor 3 from the first two terms only half of - 4 is - 2, (- 2)^{2} = 4 move - 4 out: 3 \times (- 4) = - 12 vertex (2, - 11)

Multiply by a on the way out

The

- 4

inside

3 (x^{2} - 4 x + 4 - 4)

is multiplied by the

3

in front, so when it leaves the brackets it becomes

3 \times (- 4) = - 12

, not

- 4

. Forgetting this is the most common completing-the-square error. Always check by expanding:

3 (x - 2)^{2} - 11 = 3 (x^{2} - 4 x + 4) - 11 = 3 x^{2} - 12 x + 12 - 11 = 3 x^{2} - 12 x + 1

✓.

VideoAdd and subtract the right constant to rewrite standard form as vertex form and read off the vertex.Watch on YouTube

Graphing y = ax² + bx + c: choose a method

Method	Best when	Example
Factor, then use zeros and symmetry	the expression factors with whole numbers	$y = x^{2} - 2 x - 8 = (x - 4) (x + 2)$
Complete the square, then transform y = x²	you want the vertex and the shape	$y = 3 x^{2} - 12 x + 1 = 3 (x - 2)^{2} - 11$
Partial factoring	you need the axis and vertex quickly	$y = 2 x^{2} - 8 x + 5 = 2 x (x - 4) + 5$
Graphing technology	the numbers are decimals or messy	$h = - 4.9 t^{2} + 50 t + 1.5$

Four ways to graph a quadratic in standard form

Partial factoring finds the axis without factoring completely. Take out a common factor from the $x^{2}$ and $x$ terms only: $y = 2 x^{2} - 8 x + 5 = 2 x (x - 4) + 5$ . The product $2 x (x - 4)$ is zero when $x = 0$ or $x = 4$ , and at both of those $x$ -values, $y = 0 + 5 = 5$ . So $(0, 5)$ and $(4, 5)$ are at the same height, and by symmetry the axis is halfway: $x = 2$ . Substituting gives $y = 2 (2)^{2} - 8 (2) + 5 = 8 - 16 + 5 = - 3,$ so the vertex is $(2, - 3)$ . Technology is the right tool for a model like $h = - 4.9 t^{2} + 50 t + 1.5$ , the height in metres of a model rocket launched from a 1.5 m platform. A graphing tool shows the vertex at about $(5.10, 129.1)$ : a maximum height of about $129$ m after about $5.1$ s, landing after about $10.2$ s.

Loading interactive…

Enter a, b and c in the standard form panel and open the Complete the square tab to see each step for that parabola. The graph marks the vertex, and the vertex form panel shows the result.

When fractions appear

This course completes the square only when no fractions appear: when

a = 1

and

b

is even, or when

a

divides

b

and

\frac{b}{a}

is even. For

y = x^{2} + 5 x + 1

, half of

5

\frac{5}{2}

, so the method still works but needs fractions. In that case partial factoring or technology is quicker.

Completing the Square

Key things to remember

Completing the square, step by step

Graphing y = ax² + bx + c: choose a method

Worked examples

Where marks get lost

Flashcards 8

Practice