## Completing the Square Video Lesson

## How to Complete the Square

**To complete the square for 𝑥**

^{2}+b𝑥+c:

- Divide b by 2 and write this inside brackets squared as (𝑥 +
^{b}/_{2})^{2} - Square the value of
^{b}/_{2}and subtract it. - Add the constant term of c to this result.
- The final result is given by (𝑥 +
^{b}/_{2})^{2}+ c – (^{b}/_{2})^{2}.

For example, complete the square for y = 𝑥^{2} + 4𝑥 + 1.

**Step 1. Divide the coefficient of 𝑥 by 2**

The coefficient of 𝑥 is 4.

4 ÷ 2 = 2 and so, we write (𝑥 + 2)^{2}.

This step is shown with a purple arrow.

**Step 2. Square the value found in step 1 and subtract it**

The value found in step 1 is half of the 𝑥 coefficient. Half of 4 is 2.

We square 2 and subtract this.

2^{2} = 4 and so we subtract 4.

This step is shown with the green arrow.

**Step 3. Bring down the constant term**

In the equation y = 𝑥^{2} + 4𝑥 + 1, the constant term is 1.

We add this value to our result.

This step is shown with the red arrow.

Finally, we simplify the constant terms.

-4 + 1 = -3 and so we write this as y = (𝑥 + 2)^{2} – 3.

### Formula for Completing the Square

The formula for completing the square for a quadratic, y = 𝑥^{2} + b𝑥 + c, is .

If y = 𝑥^{2} + 6𝑥 + 4, then b = 6 and c = 4.

Therefore ^{b}/_{2} = 3.

Therefore becomes .

We can simplify 4 – (3)^{2} to 4 – 9 which equals -5.

The final form of the answer is y = (𝑥 + 3)^{2} – 5.

Here is an example of completing the square when the value of b is odd.

Complete the square for y = 𝑥^{2} + 3𝑥 + 3.

When the coefficient of 𝑥 is odd, write it as a fraction over 2 when completing the square.

**Step 1. Divide the coefficient of 𝑥 by 2**

The coefficient of 𝑥 is 3.

To divide 3 by 2, write this as ^{3}/_{2}.

We write this inside the bracket as (𝑥 + ^{3}/_{2})^{2}.

This step is written in purple below.

**Step 2. Square this value and subtract it**

To square ^{3}/_{2}, square the numerator and denominator separately.

3^{2} = 9 and 2^{2} = 4.

We subtract ^{9}/_{4} from the brackets written in step 1.

This step is written in green below.

**Step 3. Bring down the constant term**

The constant term of y = 𝑥^{2} + 3𝑥 + 3 is the number on its own. It is 3.

We add 3 to the equation built so far.

This step is shown in red below.

Finally, we simplify by collecting the constant terms of –^{9}/_{4} and + 3.

3 can be written as ^{12}/_{4}.

–^{9}/_{4} + ^{12}/_{4} = ^{3}/_{4}.

Therefore the result is y = (𝑥 +^{ 3}/_{2})^{2} + ^{3}/_{4}.

## What is Completing the Square?

**Completing the square is an algebraic method used to rearrange a quadratic equation from y = a𝑥 ^{2}+b𝑥+c to the form of y = a(𝑥+b)^{2}+c. Completing the square allows us to solve quadratic equations that cannot be factorised and to find the turning point of a quadratic. It is possible to complete the square for any quadratic equation.**

For example, completing the square with y = 𝑥^{2} + 4𝑥 + 1 rearranges the equation to y = (𝑥 + 2)^{2} – 3.

y = 𝑥^{2} + 4𝑥 + 1 and y = (𝑥 + 2)^{2} – 3 are the same equation, just rearranged.

### Proof of Completing the Square

Completing the square can be proven algebraically by expanding (𝑥 + ^{b}/_{2})^{2} to get 𝑥^{2} + b𝑥 + (^{b}/_{2})^{2}. This means that any quadratic of the form 𝑥^{2} + b𝑥* *can be rearranged to (𝑥 + ^{b}/_{2})^{2} – (^{b}/_{2})^{2}.

Adding a constant term of c to both sides of the equation, any quadratic of the form 𝑥^{2} + b𝑥 + c can be written as .

### Visual Proof of Completing the Square

Completing the square can be shown visually using the following steps.

- Consider a square of area 𝑥
^{2}and a rectangle of area b𝑥. - Divide the rectangle b𝑥 in half to form two rectangles of area
^{b}/_{2}𝑥. - Place these two rectangles around the square of 𝑥
^{2}

4. The square of 𝑥^{2} plus the two rectangles almost form a square with side lengths of 𝑥 + ^{b}/_{2}. However, there is a small square of side length ^{b}/_{2} missing.

The area of the overall square (outlined in red above) would be (𝑥 + ^{b}/_{2})^{2}.

The area of the purple square above would be (^{b}/_{2})^{2}.

Therefore (𝑥 + ^{b}/_{2})^{2} – (^{b}/_{2})^{2} = 𝑥^{2} + b𝑥.

Adding a constant term of c to each side of the equation tells us that .

## How to Complete the Square with a Coefficient

**To complete the square with a coefficient of 𝑥**

^{2}:

- Factorise the coefficient
- Complete the square inside the bracket
- Simplify the terms inside the bracket Expand the bracket

For example, complete the square for y = 2𝑥^{2} + 8𝑥 – 6.

**Step 1. Factorise the coefficient**

The coefficient of 𝑥^{2} is 2.

We factorise the expression by bringing a 2 in front of the brackets and dividing every term inside the brackets by 2.

We get y = 2[𝑥^{2} + 4𝑥 – 3]

**Step 2. Complete the square inside the bracket**

Inside the bracket of y = 2[𝑥^{2} + 4𝑥 – 3] is 𝑥^{2} + 4𝑥 – 3. We complete the square with this.

i) We halve the 𝑥 coefficient of 4 to get 2. Therefore inside the bracket we have (𝑥 + 2)^{2}.

ii) We subtract this value squared. 2 squared is 4 so we get (𝑥 + 2)^{2} – 4.

iii) We bring the constant term down. The constant term is -3 so we get (𝑥 + 2)^{2} – 4 – 3.

**Step 3. Simplify the terms inside the bracket**

We can simplify (𝑥 + 2)^{2} – 4 – 3 to (𝑥 + 2)^{2} – 7.

**Step 4. Expand the bracket**

We have y = 2[ (𝑥 + 2)^{2} – 7 ].

To expand this, we multiply the (𝑥 + 2)^{2} term and the -7 term both by 2.

We get y = 2(𝑥 + 2)^{2} – 14.

### Completing the Square with a Negative Coefficient

In this example we will complete the square with a negative coefficient of 𝑥^{2}.

Complete the square for y = -3𝑥^{2} + 6𝑥 – 9. In this example, the coefficient of 𝑥^{2} is -3.

**Step 1. Factorise the coefficient of 𝑥 ^{2}**

We factorise the coefficient of -3 by writing -3 in front of the brackets and dividing each term within the brackets by -3.

Because we are factorising a negative number, the signs of the terms inside the bracket switch from + to – or – to + respectively.

y = -3𝑥^{2} + 6𝑥 – 9 becomes y = -3[ 𝑥^{2} – 2𝑥 + 3 ].

**Step 2. Complete the square inside the bracket**

Inside the bracket of y = -3[𝑥^{2} – 2𝑥 + 3] is 𝑥^{2} – 2𝑥 + 3. We complete the square with this.

i) We halve the 𝑥 coefficient of -2 to get -1. Therefore inside the bracket we have (𝑥 – 1)^{2}.

ii) We subtract this value squared. (-1) squared is 1 so we get (𝑥 – 1)^{2} – 1.

iii) We bring the constant term down. The constant term is 3 so we get (𝑥 – 1)^{2} – 1 + 3.

**Step 3. Simplify terms inside the bracket**

Inside the bracket, (𝑥 – 1)^{2} – 1 + 3 can be simplified to (𝑥 – 1)^{2} + 2.

**Step 4. Expand the bracket**

We have y = -3[ (𝑥 – 1)^{2} + 2 ].

To expand this, we multiply the (𝑥 – 1)^{2} term and the +2 term both by -3.

We get y = -3(𝑥 – 1)^{2} – 6.

## How to Solve an Equation by Completing the Square

**Any quadratic equation can be solved by completing the square. After completing the square, move the constant terms so that they are on a different side of the equation to the squared brackets to get (𝑥 + a) ^{2} = k. Then square root both sides to get 𝑥 + a = ±√k. Subtracting a from both sides gives the solution as 𝑥 = -a ±√k.**

For example, solve the equation 𝑥^{2} + 8𝑥 + 5 = 0.

**Step 1. Complete the square**

i) Halve the coefficient of 𝑥.

The coefficient of 𝑥 is 8 and half of 8 is 4. We start with writing (𝑥 + 4)^{2}.

ii) Square this value and subtract it

4 squared is 16, so we subtract 16 to get (𝑥 + 4)^{2} – 16

iii) Bring down the constant term.

The constant term of 𝑥^{2} + 8𝑥 + 5 is 5, so we add 5 to get (𝑥 + 4)^{2} – 16 + 5.

This simplifies to (𝑥 + 4)^{2} – 11.

**Step 2. Solve for 𝑥**

i) First separate the squared brackets and any constant terms on either side of the equals sign

We add 11 to both sides so that (𝑥 + 4)^{2} is on the left of the equals sign and 11 is on the right of it.

ii) Square root both sides of the equation

The square root of (𝑥 + 4)^{2} is just 𝑥 + 4.

The square root of 11 is ±√11. We need to remember to take both the positive an negative solutions.

iii) Solve for 𝑥

We have 𝑥 + 4 = ±√11.

We subtract 4 from both sides to get 𝑥 = -4 ±√11.

Here is another example of solving a quadratic equation by completing the square.

For example, solve the equation 𝑥^{2} + 6𝑥 – 1 = 0.

**Step 1. Complete the square**

i) Halve the coefficient of 𝑥.

The coefficient of 𝑥 is 6 and half of 6 is 3. We start with writing (𝑥 + 3)^{2}.

ii) Square this value and subtract it

3 squared is 9, so we subtract 9 to get (𝑥 + 3)^{2} – 9

iii) Bring down the constant term.

The constant term of 𝑥^{2} + 6𝑥 – 1 is -1, so we subtract 1 to get (𝑥 + 3)^{2} – 10.

**Step 2. Solve for 𝑥**

i) First separate the squared brackets and any constant terms on either side of the equals sign

We add 10 to both sides so that (𝑥 + 3)^{2} is on the left of the equals sign and 10 is on the right of it.

ii) Square root both sides of the equation

The square root of (𝑥 + 3)^{2} is just 𝑥 + 3.

The square root of 10 is ±√10. We need to remember to take both the positive an negative solutions.

iii) Solve for 𝑥

We have 𝑥 + 3 = ±√10.

We subtract 3 from both sides to get 𝑥 = -3 ±√10.

## Finding the Vertex by Completing the Square

**Completing the square allows a quadratic to be written in the form y = a(𝑥 + h) ^{2} + k. The vertex of the parabola is found at the coordinate (-h, k). For example, the parabola y = 2(𝑥 + 3)^{2} + 1 has a vertex at the coordinate (-3, 1). **

The vertex is the turning point of a parabola.

The graph below shows the vertex of y = 2(𝑥 + 3)^{2} + 1 at (-3, 1).

The vertex is found at the 𝑥 value that causes the brackets to be equal to zero.

In the equation y = 2(𝑥 + 3)^{2} + 1, the brackets are squared.

When we square a value, the result is always positive. Therefore the bracket squared can never be negative.

This means that the smallest value that can be output by the equation y = 2(𝑥 + 3)^{2} + 1 is when the bracket is worth zero.

The value of 𝑥 that causes this to happen is at 𝑥 = -3.

When 𝑥 = -3, the equation of y = 2(𝑥 + 3)^{2} + 1 becomes y = 1.

Therefore we find the minimum value at 𝑥 = -3, y = 1 or written as a coordinate (-3, 1).

Here is an example of finding the vertex of a quadratic by completing the square.

For example, find the vertex of y = 𝑥^{2} – 2𝑥 + 3

**Step 1. Complete the square**

i) Halve the coefficient of 𝑥.

The coefficient of 𝑥 is -2 and half of -2 is -1. We start with writing (𝑥 – 1)^{2}.

ii) Square this value and subtract it

(-1) squared is 1, so we subtract 1 to get (𝑥 – 1)^{2} – 1

iii) Bring down the constant term.

The constant term of 𝑥^{2} – 2𝑥 + 3 is 3, so we add 3 to get (𝑥 – 1)^{2} + 2.

**Step 2. Read the vertex from the complete the square form**

The 𝑥 coordinate of the vertex is the value of 𝑥 that makes the bracket equal to zero.

In y = (𝑥 – 1)^{2}** **+ 2, the expression inside the bracket is 𝑥 – 1.

This is equal to zero when 𝑥 = 1. Therefore the 𝑥 coordinate of the vertex is at 𝑥 = 1.

The y coordinate of the vertex is simply the constant term in y = (𝑥 – 1)^{2} + 2. This is just the +2 at the end.

The y coordinate of the vertex is at y = 2.

Therefore the coordinate of the vertex is at (-1 , 2).

Notice how the value of the 𝑥 coordinate is opposite sign to the sign written in the brackets but the y coordinate is the same sign as the constant term at the end.

Here are some examples of reading the turning point from an equation in complete the square form.

Equation | Coordinates of the Vertex |

y = a(𝑥 + h)^{2} + k | (-h, k) |

y = (𝑥 + 3)^{2} + 5 | (-3, 5) |

y = 5(𝑥 + 1)^{2} + 2 | (-1, 2) |

y = (𝑥 – 3)^{2} + 7 | (3, 7) |

y = 2(𝑥 – 4)^{2} – 3 | (4, -3) |

y = 0.1(𝑥 + 2)^{2} – 6 | (-2, -6) |

y = (𝑥 – 5)^{2} | (5, 0) |