Q13 Complete the square without fractional coefficients

Ontario Grade 10 Mathematics

Quadratic Relations

Rewriting Quadratics in Vertex Form — No Fractions Required

You already know how to expand a binomial like

(x + 3)^2

and get

x^2 + 6x + 9

. Completing the square reverses that process — you start with a messy trinomial such as

x^2 + 6x + 9

and rewrite it as the tidy square

(x + 3)^2

. This skill is the bridge between standard form

y = ax^2 + bx + c

and vertex form

y = a(x - h)^2 + k

, which tells you the vertex of the parabola right away. In this lesson every example is designed so that no fractions appear during the work, which keeps the focus firmly on the method itself.

What you will learn

Recognise when a quadratic expression is a perfect-square trinomial.
Rewrite a quadratic expression of the form $x^2 + bx + c$ in vertex form by completing the square.
Rewrite a quadratic expression of the form $ax^2 + bx + c$ (where $a$ divides evenly into $b$ and the process stays whole-number) in vertex form by completing the square.
Identify the vertex of a parabola directly from vertex form.
Avoid the common sign and arithmetic errors that arise during completing the square.

Grade 9 Bridge — Perfect-Square Trinomials

A perfect-square trinomial is a three-term polynomial that is the result of squaring a binomial. For example,

(x + 5)^2 = x^2 + 10x + 25

. Notice the pattern: the last term,

25

, is exactly the square of half the middle coefficient (

\frac{10}{2} = 5

, and

5^2 = 25

). This pattern is the engine of the whole method.

Check a few more:

(x + 4)^2 = x^2 + 8x + 16

. Half of

8

is

4

, and

4^2 = 16

. Also,

(x - 7)^2 = x^2 - 14x + 49

. Half of

14

is

7

, and

7^2 = 49

. The sign of the middle term tells you the sign inside the bracket, but the last term is always positive because any number squared is positive.

This means: given

x^2 + bx

, you can always figure out exactly what constant to add to turn it into a perfect-square trinomial. That constant is

\left(\frac{b}{2}\right)^2

. The lessons in this study guide focus on values of

b

that are even integers, so the division by

2

never creates a fraction.

(\frac{b}{2})^2

$(x + n)^2 = x^2 + 2nx + n^2$ — the last term is always the square of $n$ .
Given $x^2 + bx$ , the magic number to add is $\left(\frac{b}{2}\right)^2$ .
When $b$ is even, $\frac{b}{2}$ is a whole number and no fractions appear.
The sign inside the squared bracket matches the sign of the middle term.

What Does 'Complete the Square' Mean?

The phrase complete the square comes from geometry. Imagine a square with side length

x

and a rectangle attached to it with width

\frac{b}{2}

. To make a larger perfect square you need a small corner piece of area

\left(\frac{b}{2}\right)^2

. Adding that piece 'completes' the square shape — and the algebra mirrors the geometry exactly.

In algebra, completing the square means rewriting

x^2 + bx + c

as

a(x - h)^2 + k

. The form

a(x - h)^2 + k

is called vertex form because it immediately reveals the vertex

(h, k)

of the parabola.

The key constraint: whatever you add inside the expression, you must also subtract — otherwise you change the value of the expression. For example,

x^2 + 6x + 1 = (x^2 + 6x + 9) - 9 + 1

. You added

9

and subtracted

9

in the same line, so the expression is unchanged but the first three terms are now a perfect-square trinomial.

This add-and-subtract idea is the heart of the method. Every step you take is legal because you are always adding zero (since adding and subtracting the same number nets zero).

y = a(x - h)^2 + k

Vertex form is $y = a(x - h)^2 + k$ ; the vertex of the parabola is $(h, k)$ .
To complete the square, add $\left(\frac{b}{2}\right)^2$ and subtract the same amount so the expression is unchanged.
Group the first two terms, complete the square inside the group, then simplify the constants outside.

Step-by-Step Method for $x^2 + bx + c$

Here is a clear procedure you can follow every time. Start with

y = x^2 + bx + c

. Step 1 — identify

b

and compute

\left(\frac{b}{2}\right)^2

. Step 2 — rewrite the expression by adding and subtracting that number right after the

bx

term:

y = x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c

. Step 3 — group the first three terms and factor them as a perfect square:

y = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c

. Step 4 — combine the leftover constants into a single number

k

.

After those four steps the expression is in vertex form. The vertex is at

\left(-\frac{b}{2},\, k\right)

. Remember: vertex form is written as

(x - h)^2

, so if you get

(x + 3)^2

that means

h = -3

, not

h = 3

. Keep that sign rule in mind to read the vertex correctly.

When the leading coefficient

a

is not

1

, there is one extra step: factor

a

out of the first two terms only before applying the procedure above. This lesson only covers cases where that factoring stays as whole numbers — for example,

2x^2 + 8x

factors as

2(x^2 + 4x)

, and then you complete the square inside the brackets.

y = (x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c

Compute $\left(\frac{b}{2}\right)^2$ first — this is the number you add and subtract.
Add and subtract inside the same expression so you change nothing.
Factor the perfect-square trinomial, then simplify the leftover constants.
When $a \neq 1$ , factor $a$ out of the first two terms before completing the square.
Read the vertex from $(x - h)^2 + k$ carefully; the sign of $h$ flips.

Why the Vertex Form Is So Useful

Once an equation is in vertex form

y = a(x - h)^2 + k

, you can read off important information without any further calculation. The vertex

(h, k)

is the turning point of the parabola. If

a > 0

the parabola opens upward and the vertex is the minimum point. If

a < 0

it opens downward and the vertex is the maximum point.

The axis of symmetry is the vertical line

x = h

. It passes right through the vertex and divides the parabola into two mirror-image halves. Knowing the vertex and the direction of opening is enough to sketch a reasonable parabola.

These facts make vertex form a powerful tool. Whenever you need the vertex or the axis of symmetry quickly, complete the square to convert from standard form.

The vertex is $(h, k)$ — read directly from $a(x - h)^2 + k$ .
Axis of symmetry: $x = h$ .
Sign of $a$ tells you if the parabola opens up ( $a > 0$ ) or down ( $a < 0$ ).

Completing the Square — Step Comparison for Both Examples

Stage	Example 1: $x^2 - 10x + 3$	Example 2: $2x^2 + 12x - 4$
Factor out $a$ (if $a \neq 1$ )	Not needed ( $a = 1$ )	$2(x^2 + 6x) - 4$
Find magic number $\left(\frac{b}{2}\right)^2$	$(-5)^2 = 25$	$3^2 = 9$
Add and subtract magic number	$x^2 - 10x + 25 - 25 + 3$	$2(x^2 + 6x + 9 - 9) - 4$
Move subtracted term outside brackets	Already outside: $-25 + 3$	$2(x^2 + 6x + 9) - 18 - 4$
Factor trinomial and simplify	$(x-5)^2 - 22$	$2(x+3)^2 - 22$
Vertex	$(5,\ -22)$	$(-3,\ -22)$

Worked example

Example 1 — Leading Coefficient of 1

Rewrite

y = x^2 - 10x + 3

in vertex form, then state the vertex and axis of symmetry.

Identify b and find the magic number
The coefficient of $x$ is $b = -10$ . Compute $\frac{b}{2} = \frac{-10}{2} = -5$ , then square it: $(-5)^2 = 25$ . This is the number you will add and subtract.
$(\frac{-10}{2})^2 = (-5)^2 = 25$
Add and subtract 25 inside the expression
Write the original expression and insert $+25$ and $-25$ immediately after the $-10x$ term. Because you are adding and subtracting the same value, the expression has not changed.
$y = x^2 - 10x + 25 - 25 + 3$
Factor the perfect-square trinomial
The first three terms $x^2 - 10x + 25$ form a perfect-square trinomial. Factor them as $(x - 5)^2$ , since $(-5)^2 = 25$ and $2 \times (-5) = -10$ .
$y = (x - 5)^2 - 25 + 3$
Combine the constants
Simplify the two leftover numbers: $-25 + 3 = -22$ . The expression is now in vertex form.
$y = (x - 5)^2 - 22$
State the vertex and axis of symmetry
Vertex form is $y = (x - h)^2 + k$ . Here $h = 5$ and $k = -22$ , so the vertex is $(5, -22)$ . The axis of symmetry is the vertical line $x = 5$ .

Answer:

y = (x - 5)^2 - 22

; vertex

(5,\,-22)

; axis of symmetry

x = 5

.

Check: Expand

(x - 5)^2 - 22 = x^2 - 10x + 25 - 22 = x^2 - 10x + 3

. This matches the original expression, so the answer is correct.

Worked example

Example 2 — Leading Coefficient Greater Than 1

Rewrite

y = 2x^2 + 12x - 4

in vertex form, then state the vertex.

Factor the leading coefficient out of the first two terms only
The leading coefficient is $a = 2$ . Factor $2$ out of $2x^2 + 12x$ only; leave the constant $-4$ outside the brackets for now.
$y = 2(x^2 + 6x) - 4$
Find the magic number for the expression inside the brackets
Inside the brackets, $b = 6$ . Compute $\frac{6}{2} = 3$ , then $3^2 = 9$ . You will add and subtract $9$ inside the brackets.
$(\frac{6}{2})^2 = 3^2 = 9$
Add and subtract 9 inside the brackets
Insert $+9$ and $-9$ inside the brackets after the $6x$ term. The expression is still equal to the original.
$y = 2(x^2 + 6x + 9 - 9) - 4$
Separate the -9 from inside the brackets
Move the $-9$ outside the brackets. Because it was inside brackets multiplied by $2$ , it becomes $2 \times (-9) = -18$ when it comes out. The three terms that remain inside the brackets form a perfect square.
$y = 2(x^2 + 6x + 9) - 18 - 4$
Factor the trinomial and combine the constants
Factor $x^2 + 6x + 9 = (x + 3)^2$ . Then combine the outside constants: $-18 - 4 = -22$ .
$y = 2(x + 3)^2 - 22$
State the vertex
Vertex form is $y = a(x - h)^2 + k$ . Here $2(x + 3)^2 - 22 = 2(x - (-3))^2 + (-22)$ , so $h = -3$ and $k = -22$ . The vertex is $(-3, -22)$ .

Answer:

y = 2(x + 3)^2 - 22

; vertex

(-3,\,-22)

.

Check: Expand:

2(x + 3)^2 - 22 = 2(x^2 + 6x + 9) - 22 = 2x^2 + 12x + 18 - 22 = 2x^2 + 12x - 4

. This matches the original, confirming the answer is correct.

Common mistakes and how to avoid them

Forgetting to subtract the magic number after adding it, which changes the value of the expression.

Correction: Always add and subtract

\left(\frac{b}{2}\right)^2

in the same step. The net effect is adding zero, so the expression stays equal to the original.

When

a \neq 1

, not multiplying the subtracted magic number by

a

when bringing it outside the brackets.

Correction: Whatever sits inside brackets that are multiplied by

a

must be multiplied by

a

when moved outside. If you add and subtract

9

inside brackets with a factor of

2

, the term that comes out is

-18

, not

-9

.

Reading the vertex sign incorrectly from vertex form — for example, writing the vertex as

(-5, -22)

for

(x - 5)^2 - 22

.

Correction: Vertex form is

(x - h)^2 + k

. In

(x - 5)^2 - 22

, the value inside is

-5

, so

h = +5

. The vertex is

(5, -22)

.

Using an odd value of

b

(such as

b = 7

) and creating a fraction

\frac{7}{2}

without realising the question required whole numbers.

Correction: In Q13 problems,

b

is always even or

a

divides evenly into

b

, so check that your division by

2

stays a whole number. If it does not, re-read the question — you may have copied a coefficient incorrectly.

Trying to include the constant term

c

inside the brackets when factoring out

a

.

Correction: When

a \neq 1

, factor

a

out of only the first two terms. Leave the constant

c

outside the brackets and deal with it at the end when combining constants.

Lesson summary

A perfect-square trinomial is a trinomial that factors as $(x + n)^2$ . Recognising this pattern is the foundation of the whole method.
The number that completes the square is $\left(\frac{b}{2}\right)^2$ , where $b$ is the coefficient of $x$ . When $b$ is even this is always a whole number.
You add and subtract the same number so that the expression is unchanged — you are effectively adding zero.
When the leading coefficient $a$ is not $1$ , factor $a$ out of the first two terms first, complete the square inside the brackets, then multiply the subtracted term by $a$ as it comes outside.
Vertex form $y = a(x - h)^2 + k$ reveals the vertex $(h, k)$ and axis of symmetry $x = h$ directly.
Always verify your answer by expanding the vertex form — you should recover the original standard-form expression exactly.

Check your understanding

Question 1

What number must be added and subtracted to complete the square in

x^2 + 8x

?

$4$
$8$
$16$
$64$

Show answer and explanation

16

Take half of the coefficient of

x

:

\frac{8}{2} = 4

. Then square it:

4^2 = 16

. The magic number is

16

.

Question 2

Which of the following is the correct vertex form of

y = x^2 - 6x + 2

?

$y = (x - 3)^2 + 2$
$y = (x - 3)^2 - 7$
$y = (x + 3)^2 - 7$
$y = (x - 6)^2 - 34$

Show answer and explanation

y = (x - 3)^2 - 7

Half of

-6

is

-3

, and

(-3)^2 = 9

. Add and subtract

9

:

x^2 - 6x + 9 - 9 + 2 = (x-3)^2 - 7

. The vertex form is

y = (x-3)^2 - 7

.

Question 3

What is the vertex of the parabola

y = 3(x + 4)^2 - 5

?

$(4, -5)$
$(-4, 5)$
$(-4, -5)$
$(4, 5)$

Show answer and explanation

(-4, -5)

Vertex form is

a(x - h)^2 + k

. Here

3(x + 4)^2 - 5 = 3(x - (-4))^2 + (-5)

, so

h = -4

and

k = -5

. The vertex is

(-4, -5)

.

Question 4

A student rewrites

y = 2x^2 + 8x + 1

and gets

y = 2(x + 2)^2 - 7

. Is this correct?

Yes, it is fully correct.
No — the magic number was not multiplied by $2$ when brought outside the brackets.
No — the student should have factored $2$ out of all three terms.
No — the student used the wrong sign inside the bracket.

Show answer and explanation

Yes, it is fully correct.

Check by expanding:

2(x+2)^2 - 7 = 2(x^2 + 4x + 4) - 7 = 2x^2 + 8x + 8 - 7 = 2x^2 + 8x + 1

. This matches the original, so the student's answer is correct.

Key terms

Standard form: A quadratic written as $y = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are real numbers and $a \neq 0$ .
Vertex form: A quadratic written as $y = a(x - h)^2 + k$ , where the vertex of the parabola is the point $(h, k)$ .
Perfect-square trinomial: A three-term polynomial that results from squaring a binomial, such as $x^2 + 10x + 25 = (x + 5)^2$ .
Completing the square: An algebraic method that rewrites a quadratic expression by adding and subtracting the same carefully chosen constant so that part of the expression becomes a perfect-square trinomial.
Vertex: The turning point of a parabola — its highest point if the parabola opens down, or its lowest point if it opens up.
Axis of symmetry: The vertical line that passes through the vertex and divides the parabola into two mirror-image halves. Its equation is $x = h$ .
Leading coefficient: The number $a$ in front of the $x^2$ term in a quadratic expression. It controls how wide or narrow the parabola is and whether it opens up or down.

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About this lesson

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 10 Mathematics (MPM2D), study topic Q13. It is a study resource, not an official curriculum publication.

Official curriculum reference

Report a correction or ask a question

Q13 · Complete the square without fractional coefficients

Ontario Grade 10 Mathematics

Rewriting Quadratics in Vertex Form — No Fractions Required

What you will learn

Grade 9 Bridge — Perfect-Square Trinomials

What Does 'Complete the Square' Mean?

Step-by-Step Method for $x^2 + bx + c$

Why the Vertex Form Is So Useful

Completing the Square — Step Comparison for Both Examples

Example 1 — Leading Coefficient of 1

Example 2 — Leading Coefficient Greater Than 1

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through MPM2D

About this lesson