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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.

'Use Completing the Square to Solve Quadratics' worksheet

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:  

Worksheet Overview

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 you want to write the quadratic expression in the form (x ± ?)² ± ?. Here is an example to demonstrate the method. 

Complete the square on 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 and 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 it 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 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 and these are best left as surds unless you are told to convert them to decimals.

Here is another example.

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 x² then you 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.

 

 

 

 

Below you will find the 7 steps to solving the equation by completing the square.

Place each step in the correct order.

Column A

Column B

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

By filling in the spaces below complete the method for solving the equation x² + 8x - 9 = 0.

 

Column A

Column B

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

By filling in the spaces below complete the method for solving the equation x² - 2x - 9 = 0.

 

Column A

Column B

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

Below you will find the 8 steps to solving the equation 2x² - 12x + 28 = 0 by completing the square.

Place each step in the correct order.

Column A

Column B

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

By filling in the spaces below complete the method for solving the equation 3x² + 12x + 1 = 0.

 

Column A

Column B

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

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

The solutions x = 3 ±√2 are the correct solutions to which of the equations below?

 

x² - 10x + 7

x² + 12x + 25 = 0

x² - 6x + 7 = 0

x² + x - 3 = 0

Match the equations below to their solutions.

 

Column A

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 ±√8
x² + 2x - 7 = 0
x = -1/2 ±√13/2

The solutions to the equation 6x² + 24x + 5 = 0 can be written in the form p ±√q. Choose the correct values of p and q below.

From the list of 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

Below you will find the 7 steps to solving the equation by completing the square.

Place each step in the correct order.

CORRECT ANSWER

Column A

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
First, we complete the square by halving the coefficient of x and placing it in the bracket then subtracting its square after the bracket. x² - 6x - 10 = 0 (x - 3)² - (-3)² - 10 = 0 Now simplify (x - 3)² - 19 = 0 Add 19 to both sides (x - 3)² = 19 Square root x - 3 = ±√19 Add 3 x = 3 ±√19 So x = 3 +√19 or x = 3 -√19 How did you do?
  • Question 2

By filling in the spaces below complete the method for solving the equation x² + 8x - 9 = 0.

 

CORRECT ANSWER
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. We simplify so -(4)² - 9 is -25 which is the second blank. 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 the solutions are +5 - 4 = 1 and -5 - 4 = -9.
  • Question 3

By filling in the spaces below complete the method for solving the equation x² - 2x - 9 = 0.

 

CORRECT ANSWER
EDDIE SAYS
Here is the full working. 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
  • Question 4

Below you will find the 8 steps to solving the equation 2x² - 12x + 28 = 0 by completing the square.

Place each step in the correct order.

CORRECT ANSWER

Column A

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
First, have to divide the whole equation by 2 giving x² - 6x + 4 = 0 Then 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 Now simplify (x - 3)² - 5 = 0 Subtract 5 from to both sides (x - 3)² = 5 Square root x - 3 = ±√5 Add 3 x = 3 ±√5 So x = 3 +√5 or x = 3 -√5 How did you do?
  • Question 5

By filling in the spaces below complete the method for solving the equation 3x² + 12x + 1 = 0.

 

CORRECT ANSWER
EDDIE SAYS
First, we divide everything by 3 giving x² + 4x + 1/3 = 0. Then we halve the coefficient of x in the bracket and square this after the bracket giving (x + 2)² - 4 + 1/3 = 0. Simplifying (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. Choose the correct values of p and q below.

CORRECT ANSWER
EDDIE SAYS
Here are the steps 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
  • Question 7

The solutions x = 3 ±√2 are the correct solutions to which of the equations below?

 

CORRECT ANSWER
x² - 6x + 7 = 0
EDDIE SAYS
We can answer this by taking the solution x = 3 + √2 and substituting it into each equation to find the one giving the answer zero. The only one which works is x² - 6x + 7 = 0 since (3+√2)² - 6x(3+√2) + 7 = 0.
  • Question 8

Match the equations below to their solutions.

 

CORRECT ANSWER

Column A

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
Let's look at each equation. 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.
  • Question 9

The solutions to the equation 6x² + 24x + 5 = 0 can be written in the form p ±√q. Choose the correct values of p and q below.

CORRECT ANSWER
EDDIE SAYS
Here are the steps 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) How did you do?
  • Question 10

From the list of options below select the two correct solutions to the equation 

2x² - 5x - 1 = 0

CORRECT ANSWER
5/4 +√33/4
5/4 -√33/4
EDDIE SAYS
Here's the correct solution 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 done now.
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