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Solve Complex Quadratic Equations

In this worksheet, students will learn how to solve a quadratic equation of the form ax² + bx + c = 0 by factorising (including equations which require rearranging).

'Solve Complex Quadratic Equations' 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

In this activity, we will be looking at more complex quadratic equations, which can be solved by factorising.

 

The method of factorising a quadratic expression in the form ax² + bx + c into two brackets is covered in another activity.

If you haven't completed this yet or are not feeling totally confident with this skill, it would be best to review this before continuing here. 

 

 

e.g. 2x² + x - 3 = 0

 

So, the equation 2x² + x - 3 = 0 can be factorised into two brackets in the form: (x - 1)(2x + 3)

 

Now, remember that each our brackets will be multiplied together to give the answer 0.

If I tell you that I have multiplied together two numbers and the answer is zero, can you tell me anything about my two numbers?

Well, you should be able to tell me that at least one of my numbers is zero itself.

 

Going back to our equation, this means that either: (x - 1) = 0 or (2x + 3) = 0 ​(They cannot both be zero.) 

If x - 1 = 0 then x = 1 is a solution.

Alternatively, if 2x + 3 = 0 then x = -3/2 (-1.5) is a solution.

 

So there are two possible solutions to this quadratic equation: x = 1 and x = -1.5

It is a feature of quadratic equations that they nearly always have two solutions.

 

 

Here's another example, this time we'll just include the lines of working you would need to show:

6n² - 5n - 4 = 0

(2n + 1)(3n - 4) = 0

Either 2n + 1 = 0 or 3n - 4 = 0

So n = -1/2 or n = 4/3

 

 

Finally, we'll look at one more example:

4z(z - 1) = 3

 

Now, this equation looks like it has already been factorised.

However, you will notice the right-hand side is 3, not 0.

Quadratic equations can only be solved if the right-hand side is zero.

 

So we need to rearrange the equation first and this means we must expand the brackets as our starting point:

4z² - 4z = 3

 

Now subtract 3 from both sides:

4z² - 4z - 3 = 0

 

Now our equation is in the correct form for us to factorise into two brackets and then solve:

 4z² - 4z - 3 = 0

(2z - 3)(2z + 1) = 0

Either 2z - 3 = 0 or 2z + 1 = 0

So z = 3/2 or z = -1/2

 

 

 

Ready to try some for yourself now?

 

In this activity, we will solve quadratic equations by ensuring that they conform to the style of ax² + bx + c = 0 first, then factorising them to find possible values for our variables. 

Consider this quadratic equation:

 

6x² - 13x + 5 = 0

 

In the list below, you will see the steps to solving this equation. 

 

Put these steps into the correct order by matching each stage to its correct position.

Column A

Column B

1st
x = 1/2 or x = 5/3
2nd
2x - 1 = 0 or 3x - 5 = 0
3rd
(2x - 1)(3x - 5) = 0
4th
6x² - 13x + 5 = 0

Consider this quadratic equation:

 

2x² - 5x - 3 = 0

 

Below you will see the steps for solving the equation.

 

Fill in the spaces in each line to complete the solution for this equation.

Column A

Column B

1st
x = 1/2 or x = 5/3
2nd
2x - 1 = 0 or 3x - 5 = 0
3rd
(2x - 1)(3x - 5) = 0
4th
6x² - 13x + 5 = 0

Solve this equation:

 

3a² + 13a + 4 = 0

 

Choose the correct pair of solutions from the options below.

a = 1/3 or a = 4

a = 4/3 or a = 1

a = -1/3 or a = -4

a = -4/3 or a = -1

Solve this equation:

 

6y² + 11y - 10 = 0

 

Choose the correct two solutions from the options below.

y = 1

y = -5/3

y = 2/3

y = 5/2

y = -2/3

y = -5/2

Solve this equation:

 

20x² + 21x + 4 = 0

 

Choose the correct two solutions from the options below.

 

x = -1/5

x = 1/5

x = -4/5

x = 4/5

x = -1/4

x = 1/4

Consider this quadratic equation:

 

4h² - 8h = 5

 

Below you will see the steps for solving this equation.

 

Fill in the spaces in each line to find the solution for this equation.

x = -1/5

x = 1/5

x = -4/5

x = 4/5

x = -1/4

x = 1/4

The following quadratic equation only has one solution:

 

16q² + 25 = 40q

 

Choose the correct single solution from the options listed below. 

4/5

-4/5

5/4

-5/4

For each equation below, select its two correct solutions from the options given.

 x = 5/3x = -5/3x = 0x = 4/3x = 4/5
5x² = 4x
9x² - 27x + 20 = 0
9x² - 25 = 0

Solve this equation:

 

8y² = -19y - 6

 

Fill your solutions into the spaces below.

 x = 5/3x = -5/3x = 0x = 4/3x = 4/5
5x² = 4x
9x² - 27x + 20 = 0
9x² - 25 = 0

Solve this equation:

 

(3x + 2)(2x + 1) = 1

 

Type your solutions into the spaces below.

 x = 5/3x = -5/3x = 0x = 4/3x = 4/5
5x² = 4x
9x² - 27x + 20 = 0
9x² - 25 = 0
  • Question 1

Consider this quadratic equation:

 

6x² - 13x + 5 = 0

 

In the list below, you will see the steps to solving this equation. 

 

Put these steps into the correct order by matching each stage to its correct position.

CORRECT ANSWER

Column A

Column B

1st
6x² - 13x + 5 = 0
2nd
(2x - 1)(3x - 5) = 0
3rd
2x - 1 = 0 or 3x - 5 = 0
4th
x = 1/2 or x = 5/3
EDDIE SAYS
Our equation already has an answer of 0 and is in the form ax² + bx + c = 0 so we can get cracking on solving it straight away! 1) Let's start with our original equation: 6x² - 13x + 5 = 0 2) We need to factorise it: (2x - 1)(3x - 5) = 0 3) Then we need to equate each bracket to zero: 2x - 1 = 0 or 3x - 5 = 0 4) Finally let's write down the possible values of x: x = 1/2 or x = 5/3 Did you get those steps in the correct order?
  • Question 2

Consider this quadratic equation:

 

2x² - 5x - 3 = 0

 

Below you will see the steps for solving the equation.

 

Fill in the spaces in each line to complete the solution for this equation.

CORRECT ANSWER
EDDIE SAYS
These are the steps for solving this equation: 2x² - 5x - 3 = 0 (2x + 1)( x - 3) = 0 Either 2x + 1 = 0 or x - 3 = 0 So x = -1/2 or x = 3 Did you get these all correct?
  • Question 3

Solve this equation:

 

3a² + 13a + 4 = 0

 

Choose the correct pair of solutions from the options below.

CORRECT ANSWER
a = -1/3 or a = -4
EDDIE SAYS
Here are the steps for solving this equation: 3a² + 13a + 4 = 0 (3a + 1)(a + 4) = 0 3a + 1 = 0 or a + 4 = 0 a = -1/3 or a = -4 It is easy to make the mistake of getting the 1 and 4 in the wrong brackets and factorising to (3a + 4)(a + 1). However, if we expand this option we will get 7a as the middle term not 13a.
  • Question 4

Solve this equation:

 

6y² + 11y - 10 = 0

 

Choose the correct two solutions from the options below.

CORRECT ANSWER
y = 2/3
y = -5/2
EDDIE SAYS
Hopefully you spotted that you needed to choose two solutions from these choices, but which ones? Here are the steps for solving this equation: 6y² + 11y - 10 = 0 (3y - 2)(2y + 5) = 0 3y - 2 = 0 or 2y + 5 = 0 y = 2/3 or y = -5/2 How are you doing with these quadratic equations? Are you getting the hang of them?
  • Question 5

Solve this equation:

 

20x² + 21x + 4 = 0

 

Choose the correct two solutions from the options below.

 

CORRECT ANSWER
x = -4/5
x = -1/4
EDDIE SAYS
Hopefully you chose two options again here! Here are the steps for solving this equation: 20x² + 21x + 4 = 0 (5x + 4)(4x + 1) = 0 5x + 4 = 0 or 4x + 1 = 0 x = -4/5 or x = -1/4 How did you do with this one?
  • Question 6

Consider this quadratic equation:

 

4h² - 8h = 5

 

Below you will see the steps for solving this equation.

 

Fill in the spaces in each line to find the solution for this equation.

CORRECT ANSWER
EDDIE SAYS
This quadratic equation did not have an answer of 0 so we need to rearrange this before we can factorise. Here is the full working for this one: 4h² - 8h = 5 4h² - 8h - 5 = 5 - 5 4h² - 8h - 5 = 0 Now we can follow our usual steps and factorise: (2h - 5)(2h + 1) = 0 2h - 5 = 0 or 2h + 1 = 0 h = 5/2 or h = -1/2 Did you fill in those gaps accurately and in the correct order?
  • Question 7

The following quadratic equation only has one solution:

 

16q² + 25 = 40q

 

Choose the correct single solution from the options listed below. 

CORRECT ANSWER
5/4
EDDIE SAYS
We need to rearrange this into equation to give an answer of 0 before we start. We can do this by subtracting 40q from both sides: 16q² + 25 = 40q 16q² + 25 - 40q = 40q - 40q 16q² - 40q + 25 = 0 Now we can factorise and solve as usual: (4q - 5)(4q - 5) = 0 We can see that both numbers in the brackets are the same (-5), which means there is only one solution on this occasion: 4q - 5 = 0 q = 5/4 Does that make sense?
  • Question 8

For each equation below, select its two correct solutions from the options given.

CORRECT ANSWER
 x = 5/3x = -5/3x = 0x = 4/3x = 4/5
5x² = 4x
9x² - 27x + 20 = 0
9x² - 25 = 0
EDDIE SAYS
Let's work through each equation one at a time. We need to rearrange the first one so that it gives an answer of 0 before we start. Here's the working for each equation: 5x² = 4x → 5x² - 4x = 0 → x(5x - 4) = 0 → x = 0 or x = 4/5 9x² - 27x + 20 = 0 → (3x - 5)(3x - 4) = 0 → x = 5/3 or x = 4/3 9x² - 25 = 0 → (3x - 5)(3x + 5)= 0 → x = 5/3 or x = -5/3 Did you match those equations and solutions accurately?
  • Question 9

Solve this equation:

 

8y² = -19y - 6

 

Fill your solutions into the spaces below.

CORRECT ANSWER
EDDIE SAYS
We needed to do some rearranging at the start to get our equation into the format: ay² + by + c = 0 Here's the full solution to this equation: 8y² = -19y - 6 8y² + 19y + 6 = 0 (Rearranging) (8x + 3)(x + 2) = 0 8x + 3 = 0 or x + 2 = 0 x = -3/8 or x = -2 Just one more challenge to go now!
  • Question 10

Solve this equation:

 

(3x + 2)(2x + 1) = 1

 

Type your solutions into the spaces below.

CORRECT ANSWER
EDDIE SAYS
This was a tricky final question, as we had a lot of work to do! Here's the full solution: (3x + 2)(2x + 1) = 1 Expand the brackets: 6x² + 7x + 2 = 1 Subtract 1 from each side: 6x² + 7x + 1 = 0 Factorise: (x + 1)(6x + 1) = 0 x + 1 = 0 or 6x + 1 = 0 x = -1 or x = -1/6 Great work - you've completed this entire activity! Why not move on to look at more quadratic equations now and different ways of solving them (such as completing the square)?
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