 # Use Completing the Square to Solve Quadratics

In this worksheet, students will learn how to solve a quadratic equation using the method of completing the square. Key stage:  KS 4

GCSE Subjects:   Maths

GCSE Boards:   Pearson Edexcel, OCR, Eduqas, AQA

Curriculum topic:   Algebra

Curriculum subtopic:   Solving Equations and Inequalities, Algebraic Equations

Difficulty level:   ### QUESTION 1 of 10

If we take a quadratic expression, such as x² + 10x + 25, and write it in the form (x + 5)², we have completed the square since (x + 5)² is known as a perfect square.

When completing the square, we need to write the quadratic expression in the form (x ± ?)² ± ?

Here is an example to demonstrate this method.

Complete the square for the quadratic expression: x² + 8x + 3

First, we halve the coefficient of x (+ 4) and put this in the bracket.

So the perfect square is: (x + 4)²

Next, we subtract the square of the number in the bracket and replace the + 3 giving:

(x + 4)² - 4² + 3

Finally, we simplify the expression.

So x² + 8x + 3 = (x + 4)² - 13

So, I hear you ask, "How does this help with solving quadratic equations?"

When trying to solve quadratic equations which will not factorise, completing the square provides us with an alternative method which does not require a calculator, unlike using the quadratic formula.

If the expression above was the equation x² + 8x + 3 = 0, completing the square enables us to solve this using the balancing method as follows:

x² + 8x + 3 = 0

This is the same as: (x + 4)² - 13 = 0

Add 13 to both sides: (x + 4)² = 13

Take the square root of both sides: x + 4 = ±√13

(Don't forget there are two possible square roots: one positive and one negative.)

Subtract 4 from both sides: x = ±√13 - 4

So there are two solutions: x = +√13 - 4 or x = -√13 - 4

These are best left as surds unless you are told to convert them to decimals.

Here is another example to follow through.

Solve the equation x² - 5x - 4 = 0 by completing the square.

Halve the coefficient of x and subtract this value squared: (x - 5/2)² - (5/2)² - 4 = 0

Simplify: (x - 5/2)² - 41/4 = 0

Add 41/4 to both sides: (x - 5/2)² = 41/4

Square root: x - 5/2 = ±√(41/4)

Add 5/2 to both sides: x = 5/2 ±√(41/4) or x = (5 ± √41)/2

So x = (5 + √41)/2 or x = (5 - √41)/2

If there is a coefficient of , then we must divide the whole equation by this first and then continue as before.

By completing the square, solve: 3x² - 12x + 4 = 0

Divide both sides by 3: x² - 4x + 4/3 = 0

Complete the square: (x - 2)² - (-2)² + 4/3 = 0

Simplify: (x - 2)² - 8/3 = 0

Solve: (x - 2)² = 8/3

x - 2 = ±√(8/3)

x = 2 ±√(8/3)

So x = 2 + √(8/3) or x = 2 - √(8/3)

Now it's over to you to solve some quadratic equations using the method of completing the square.

These methods can be long and tricky to remember, so it is a good idea to have a pen and paper handy.

If you write out your method for each question, you can compare your working with what our maths teacher has written.

x² - 6x - 10 = 0

Below you will find the 7 steps to solving this equation using the method of completing the square.

Place each step in the correct order by matching each to its position in the process.

## Column B

1st
x = 3 ±√19
2nd
x = 3 +√19 or x = 3 -√19
3rd
(x - 3)² - 19 = 0
4th
x - 3 = ±√19
5th
(x - 3)² = 19
6th
x² - 6x - 10 = 0
7th
(x - 3)² - (-3)² - 10 = 0

x² + 8x - 9 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square

## Column B

1st
x = 3 ±√19
2nd
x = 3 +√19 or x = 3 -√19
3rd
(x - 3)² - 19 = 0
4th
x - 3 = ±√19
5th
(x - 3)² = 19
6th
x² - 6x - 10 = 0
7th
(x - 3)² - (-3)² - 10 = 0

x² - 2x - 9 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square.

## Column B

1st
x = 3 ±√19
2nd
x = 3 +√19 or x = 3 -√19
3rd
(x - 3)² - 19 = 0
4th
x - 3 = ±√19
5th
(x - 3)² = 19
6th
x² - 6x - 10 = 0
7th
(x - 3)² - (-3)² - 10 = 0

2x² - 12x + 28 = 0

Below you will find the 8 steps to solving this equation using the method of completing the square.

Place each step in the correct order by matching each to its position in the process.

## Column B

1st
2x² - 12x + 8 = 0
2nd
x = 3 ±√5
3rd
x² - 6x + 4 = 0
4th
x = 3 +√5 or x = 3 -√5
5th
(x - 3)² = 5
6th
(x - 3)² - 5 = 0
7th
(x - 3)² - (-3)² + 4 = 0
8th
x - 3 = ±√5

3x² + 12x + 1 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square.

If you need to use a fraction or division sum, write this in the form a/b using the / key to show your fractions bar.

## Column B

1st
2x² - 12x + 8 = 0
2nd
x = 3 ±√5
3rd
x² - 6x + 4 = 0
4th
x = 3 +√5 or x = 3 -√5
5th
(x - 3)² = 5
6th
(x - 3)² - 5 = 0
7th
(x - 3)² - (-3)² + 4 = 0
8th
x - 3 = ±√5

The solutions to the equation x² + 4x + 1 = 0 can be written in the form p ±√q, where p and q are integers.

Solve this equation independently then choose the correct values for p and q in the table below.

The solution x = 3 ±√2 is the correct solution to which of the equations below?

x² - 10x + 7

x² + 12x + 25 = 0

x² - 6x + 7 = 0

x² + x - 3 = 0

Match each equation below to its correct solution.

## Column B

x² + 11x + 1 = 0
x = -1 ±√8
x² + 7x - 3 = 0
x = -7/2 ±√61/2
x² + x - 3 = 0
x = -1/2 ±√13/2
x² + 2x - 7 = 0
x = -11/2 ±√117/2

The solutions to the equation 6x² + 24x + 5 = 0 can be written in the form p ±√q, where p and q are integers.

Solve this equation independently then choose the correct values for p and q in the table below.

From the options below, select the two correct solutions to the equation:

2x² - 5x - 1 = 0

5/4 +√17/4

5/4 +√33/4

5/2 -√27/2

5/4 -√17/4

5/2 +√27/2

5/4 -√33/4

• Question 1

x² - 6x - 10 = 0

Below you will find the 7 steps to solving this equation using the method of completing the square.

Place each step in the correct order by matching each to its position in the process.

## Column B

1st
x² - 6x - 10 = 0
2nd
(x - 3)² - (-3)² - 10 = 0
3rd
(x - 3)² - 19 = 0
4th
(x - 3)² = 19
5th
x - 3 = ±√19
6th
x = 3 ±√19
7th
x = 3 +√19 or x = 3 -√19
EDDIE SAYS
1) Let's start with the question: x² - 6x - 10 = 0 2) We start to complete the square by halving the coefficient of x and placing it in the bracket then subtracting its square after the bracket. (x - 3)² - (-3)² - 10 = 0 3) Now simplify: (x - 3)² - 19 = 0 4) Add 19 to both sides: (x - 3)² = 19 5) Find the square roots: x - 3 = ±√19 6) Add 3: x = 3 ±√19 7) Express the two solutions using expected notation: x = 3 +√19 or x = 3 -√19 How did you get on there? Review the Introduction before moving onto tackle the rest of this activity if you need to.
• Question 2

x² + 8x - 9 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square

EDDIE SAYS
Starting with x² + 8x - 9 = 0, we halve the coefficient of x inside the bracket and subtract this after the bracket, so the first blank is 4. Then we simplify to: (x + 4)² - 25 = 0 For the next step, the 25 is moved to the other side and then square rooted for the next. Then 4 is subtracted from each side. Finally, we can write our two solutions. Since √25 is 5, our solutions are: +5 - 4 = 1 and -5 - 4 = -9
• Question 3

x² - 2x - 9 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square.

EDDIE SAYS
Phew, what a lot of blanks to fill here! Just take them one at a time and don't worry if you make a slip. Here is the full working for this question: x² - 2x - 9 = 0 (x - 1)² - (-1)² - 9 = 0 (x - 1)² - 10 = 0 (x - 1)² = 10 x - 1 = ±√10 x = 1 ±√10 So x = 1 + √10 or x = 1 - √10 How did you get on with that one?
• Question 4

2x² - 12x + 28 = 0

Below you will find the 8 steps to solving this equation using the method of completing the square.

Place each step in the correct order by matching each to its position in the process.

## Column B

1st
2x² - 12x + 8 = 0
2nd
x² - 6x + 4 = 0
3rd
(x - 3)² - (-3)² + 4 = 0
4th
(x - 3)² - 5 = 0
5th
(x - 3)² = 5
6th
x - 3 = ±√5
7th
x = 3 ±√5
8th
x = 3 +√5 or x = 3 -√5
EDDIE SAYS
1) Let's start with our initial quadratic equation: 2x² - 12x + 8 = 0 2) Then we have to divide the whole equation by 2: x² - 6x + 4 = 0 3) Next we complete the square by halving the coefficient of x and placing it in the bracket then subtracting its square after the bracket: (x - 3)² - (-3)² + 4 = 0 4) Now let's simplify this: (x - 3)² - 5 = 0 5) Subtract 5 from to both sides (x - 3)² = 5 6) Square root both sides: x - 3 = ±√5 7) Add 3: x = 3 ±√5 8) Express our final two solutions in expected notation: So x = 3 +√5 or x = 3 -√5
• Question 5

3x² + 12x + 1 = 0

Fill in the spaces below to complete the method for solving this equation by completing the square.

If you need to use a fraction or division sum, write this in the form a/b using the / key to show your fractions bar.

EDDIE SAYS
First, we divide everything by 3: x² + 4x + 1/3 = 0 Then we halve the coefficient of x in the bracket and square this after the bracket: (x + 2)² - 4 + 1/3 = 0 Simplify: (x + 2)² - 15/3 = 0 Finally solving: (x + 2)² = 15/3 x + 2 = ±√(15/3) x = ±√(15/3) - 2 x = √(15/3) - 2 or x = -√(15/3) - 2
• Question 6

The solutions to the equation x² + 4x + 1 = 0 can be written in the form p ±√q, where p and q are integers.

Solve this equation independently then choose the correct values for p and q in the table below.

EDDIE SAYS
Here are the steps to follow to solve this one: x² + 4x + 1 = 0 (x + 2)² - 2² + 1 = 0 (x + 2)² - 3 = 0 (x + 2)² = 3 x + 2 = ±√3 x = -2 ±√3 Were you able to solve this independently? Don't forget to write out your method using a pen and paper, so you can compare it to what I have written here.
• Question 7

The solution x = 3 ±√2 is the correct solution to which of the equations below?

x² - 6x + 7 = 0
EDDIE SAYS
We can answer this by taking the solution x = 3 + √2 and substituting it into each equation. If this is a viable solution to the equation, the answer we reach will be 0. If we do not reach 0, it is not a viable solution. The answer is: x² - 6x + 7 = 0 since (3+√2)² - 6x(3+√2) + 7 = 0 All the other options give answers which are not 0. Does that make sense?
• Question 8

Match each equation below to its correct solution.

## Column B

x² + 11x + 1 = 0
x = -11/2 ±√117/2
x² + 7x - 3 = 0
x = -7/2 ±√61/2
x² + x - 3 = 0
x = -1/2 ±√13/2
x² + 2x - 7 = 0
x = -1 ±√8
EDDIE SAYS
There were lots of quadratic equations to solve here! Take them one at a time, and write our your method for each. Let's look at each equation in turn: x² + 11x + 1 = 0 → (x + 11/2)² - 117/4 = 0 → x = -11/2 ±√117/2 x² + 7x - 3 = 0 → (x + 7/2)² - 61/4 = 0 → x = -7/2 ±√61/2 x² + x - 3 = 0 → (x + 1/2)² - 13/4 = 0 → x = -1/2 ±√13/2 x² + 2x - 7 = 0 → (x + 1)² - 8 = 0 → x = -1 ±√8 And there you have it - did you match those accurately?
• Question 9

The solutions to the equation 6x² + 24x + 5 = 0 can be written in the form p ±√q, where p and q are integers.

Solve this equation independently then choose the correct values for p and q in the table below.

EDDIE SAYS
Here are the steps to solve to follow to solve this one: 6x² + 24x + 5 = 0 x² + 4x + 5/6 = 0 (x + 2)² - 2² + 5/6 = 0 (x + 2)² - 19/6 = 0 (x + 2)² = 19/6 x + 2 = ±√(19/6) x = -2 ±√(19/6) Did you get there without making any slips in your method?
• Question 10

From the options below, select the two correct solutions to the equation:

2x² - 5x - 1 = 0

5/4 +√33/4
5/4 -√33/4
EDDIE SAYS
Here's the correct solution to this question: 2x² - 5x - 1 = 0 x² - 5/2x - 1/2 = 0 (x - 5/4)² - (5/4)² - 1/2 = 0 (x - 5/4)² - 33/16 = 0 (x - 5/4)² = 33/16 x - 5/4 = ±√33/4 x = 5/4 ±√33/4 There you go - all challenges completed! Solving complex quadratics can be challenging and there are many steps to get correct in a row. This is why examiners award marks for workings, so you can still get some valuable marks even if you make a slip - remember this!
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