 Apply More 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). Key stage:  KS 4

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Difficulty level:   QUESTION 1 of 10

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

2x² + x - 3 = 0

The method of factorising a quadratic expression of the form ax² + bx + c into two brackets is covered in another activity, and here we assume you have completed that activity, but if you haven't it would be best if you do that before continuing here.

So, the equation 2x² + x - 3 = 0 can be factorised into two brackets as (x - 1)(2x + 3) = 0. Now, remember that each set of brackets stand for a number and these are multiplied together to give the answer zero. So, 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 my equation, that means that either (x - 1) = 0 or (2x + 3) = 0 (they cannot both be zero). But if x - 1 = 0 then x = 1 is a solution. Alternatively, if 2x + 3 = 0 then x = -3/2 (-1.5) is another 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.

4z² - 4z = 3

Now subtract 3 from both sides.

4z² - 4z -3 = 0

Now it 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? Have a go at the questions.

Below you will see the steps to solving the quadratic equation

6x² - 13x + 5 = 0

Match the steps to the correct order.

Column B

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

Below you see the steps for solving the equation

2x² - 5x - 3 = 0

Fill in the spaces in each line to make the solution complete.

Column B

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

Solve the equation

3a² + 13a + 4 = 0

Choose the correct pair of solutions from those given 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

From the choices given below, select the two correct solutions to the equation

6y² + 11y - 10 = 0.

y = 1

y = -5/3

y = 2/3

y = 5/2

y = -2/3

y = -5/2

Which of the solutions below are correct for the equation
20x² + 21x + 4 = 0

x = -1/5

x = 1/5

x = -4/5

x = 4/5

x = -1/4

x = 1/4

Fill in the spaces in the working below for solving the equation

4h² - 8h = 5

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 solution from the following options.

4/5

-4/5

5/4

-5/4

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

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

Solve the equation

8y² = -19y - 6

Fill your solutions in the spaces below.

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

Can you solve the following equation?

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

Fill in the spaces below with your solution.

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

Below you will see the steps to solving the quadratic equation

6x² - 13x + 5 = 0

Match the steps to the correct order.

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
To solve the equation 6x² - 13x + 5 = 0 first we factorise: (2x - 1)(3x - 5) = 0 then equate each bracket to zero: 2x - 1 = 0 or 3x - 5 = 0 finally write down the values of x: x = 1/2 or x = 5/3
• Question 2

Below you see the steps for solving the equation

2x² - 5x - 3 = 0

Fill in the spaces in each line to make the solution complete.

EDDIE SAYS
These are the steps for solving the 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 them all correct?
• Question 3

Solve the equation

3a² + 13a + 4 = 0

Choose the correct pair of solutions from those given below.

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), but if you expand these you will get 7a as the middle term, not 13a.
• Question 4

From the choices given below, select the two correct solutions to the equation

6y² + 11y - 10 = 0.

y = 2/3
y = -5/2
EDDIE SAYS
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? Getting the hang of them?
• Question 5

Which of the solutions below are correct for the equation
20x² + 21x + 4 = 0

x = -4/5
x = -1/4
EDDIE SAYS
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

Fill in the spaces in the working below for solving the equation

4h² - 8h = 5

EDDIE SAYS
Here is the full working for this one. 4h² - 8h = 5 -5 -5 4h² - 8h - 5 = 0 (2h - 5)(2h + 1) = 0 2h - 5 = 0 or 2h + 1 = 0 h = 5/2 or h = -1/2
• Question 7

The following quadratic equation only has one solution.

16q² + 25 = 40q

Choose the solution from the following options.

5/4
EDDIE SAYS
Here is the full solution to this one. 16q² + 25 = 40q -40q -40q 16q² -40q + 25 = 0 (4q - 5)(4q - 5) = 0 4q - 5 = 0 q = 5/4
• Question 8

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

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

Solve the equation

8y² = -19y - 6

Fill your solutions in the spaces below.

EDDIE SAYS
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 How's it going? Just one more left now.
• Question 10

Can you solve the following equation?

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

Fill in the spaces below with your solution.

EDDIE SAYS
Last one now. There is some work to do here. Here's the solution. (3x + 2)(2x + 1) = 1 6x² + 7x + 2 = 1 (expand) 6x² + 7x + 1 = 0 (-1 from each side) (x + 1)(6x + 1) = 0 (factorise) x + 1 = 0 or 6x + 1 = 0 x = -1 or x = -1/6 That's the final one. You're done now.
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