### Quadratic Equations

It has been my experience that students overcomplicate the topic of quadratic equations.

There are a number of different ways to solve quadratic equations in addition to factorising, such as: the ‘difference of two squares’ (Level 6); the quadratic formula (Level 7), plotting the graphs of quadratic functions (Levels 5 – 7) and completing the square (Level 8). However, in this blog I am focusing purely on factorising.

My advice is to get familiar with what a quadratic expression looks like and the steps I have shown below. There are things you need to be familiar with first and they are covered in the next section. Most of the errors I come across are where the student has either missed or got muddled with the rules concerning negative values or made a mistake rearranging the equation.

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### What you need to know first when tackling QUADRATIC EQUATIONS:

- Order of operations for solving mathematical problems
**BIDMAS**(including inverse relationships):**B**rackets**I**ndices**D**ivide and /or**M**ultiply**A**dd and / or**S**ubtract

- How to apply the four operations in fractions (i.e. addition, subtraction, multiplication and division)
- How to rearrange and solve equations
- How to expand brackets and simplify algebraic expressions
- Express problems / situations using algebraic notation
- How to factorise simple expressions (‘putting the brackets in’)

** **

#### Understanding Simultaneous Equations

### What do we mean by a quadratic expression?

A** quadratic expression** is one where the highest power of x in the expressions is x squared (x^{2}). Quadratic equations can be written in the form of:

**a**x^{2} + **b**x + **c** = 0

where **a**, **b** and **c** are real numbers and **a **is not equal to zero. This means that the values of **a**, **b **and c can be positive as well as negative. They can also be decimals or fractions.

The solutions to a quadratic equation are known as **roots **and there are generally two to find.

**Note: The roots are also the points on a graph where the quadratic curve cuts the x-axis. **

** **

### How to factorise quadratic equations

**Step 1: Rearrange (if necessary)**

Make sure the quadratic equation in the form **a**x^{2} + **b**x + **c** = 0 when factorising.

This may mean you need to rearrange the equation you are given before you start.

*Example: X ^{2} +2X = 15 rearrange to X^{2} + 2X – 15 = 0*

**Step 2: Prepare Brackets**

Note the values of **a**, **b** and **c**. and write down the initial brackets (as factorising is putting the brackets in).

*Example: In the expression X ^{2 }+ 2X – 15 = 0 a = 1, b = 2, c = -15. As a is = 1, the brackets will be:*

*(X )(X ) = 0*

**Step 3: Find the values**

When **a **is equal to 1 (as in this example), we can use the following method:

X^{2} + ( **j**+**k** )X + ( **j**x**k** )=0

where **j** and **k **are two real numbers.

The middle term (“**b**”) is the **sum** of **j** and **k**.

The third term (“**c**”) is the **product** of **j** multiplied by **k**

*Example: X ^{2} + 2X – 15 = 0 where a = 1, b = 2 and c = -15 *

* j + k = 2*

*j x k = -15*

*so…….. 5 x -3 = -15 and 5 + -3 = 2*

*Therefore the factorised quadratic expression is ( X + 5 )( X – 3) = 0*

**Step 4: Check**

It is important to check your answers are correct by expanding the brackets and simplifying the equation to make sure you get the original expression.

*Example: (X + 5 )( X – 3 ) = X ^{2 }+ 5X – 3X – 15*

* = X ^{2} + 2X – 15 *

**Step 5: Solve**

By setting each bracket to zero

**Example: **

**( X + 5 ) = 0 so X = -5**

**( X – 3 ) = 0 so X = 3**

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### What to watch out for:

**This is likely to be the method used in a non-calculator paper****When c = 0, the quadratic will look like this: X**^{2}+ bX = 0

*Example: X ^{2} – 10X = 0 which will factorise to X(X-10) = 0*

*Solutions are X=0 and X=10*

**When b=0 !!!!!****Take care where the coefficient of X**^{2}is greater than 1 (i.e. 3X^{2}) where a is bigger than 1

*Example: Factorise 3X ^{2} + 32X = – 20 *

*Step 1: Rearrange to 3X ^{2} + 32X + 20 = 0*

*Step 2: Prepare brackets. a = 3, b = 32, c = 20 so……….(3X )(X ) = 0*

*Step 3: Find the values: *

* j x k = 20 and j + 3k = 32 so………2 x 10 = 20 and 2 + 3×10 = 32*

*( 3X + 2 )( X + 10 ) = 0*

*Step 4: Check*

*( 3X + 2 )( X + 10 ) = 3X ^{2} + 30X + 2X + 20 *

* = 3X ^{2} + 32X + 20*

*Step 4: Solve*

*( 3X + 2 ) = 0 , X = -2/3*

*( X + 10 ) = 0, X = -10*

* *

### What are the examiners looking for?

- Evidence that a correct method has been used – so show your working out!!!
- The correct values in the brackets
- The final solutions for the two values of X (so don’t forget to state them clearly at the end).

**A Bit About Karen **

Karen is an experienced, PGCE qualified teacher of Mathematics at Key Stage 3 and above, including the International GCSE. She has been an Examiner and Moderator for Edexcel and therefore has good insight into what examiners are looking for and with exam techniques.

You can contact Karen about tutoring here

Karen is a skilled creative thinker and problem solver with a passion for identifying untapped talent in individuals so they are able to make a lasting transformational change into the best version of themselves. She provides a safe environment that is non-judgemental, encouraging, positive and challenging, with the freedom to experiment.

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