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Solving Quadratic Equations by Factorising Simplified Revision Notes

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Solving Quadratic Equations by Factorising

What is a Quadratic Equation?

A quadratic equation is a type of equation that looks like this: ax2+bx+c=0ax^2 + bx + c = 0 where:

  • x2x^2 means xx is multiplied by itself (so it's a "squared" term).
  • aa, bb, and cc are numbers that can be positive, negative, or zero.
  • xx is the variable we're trying to solve for (find the value of). Quadratic equations often appear when dealing with areas, paths of objects in the air, or other real-world scenarios.

Why Do We Factorise Quadratic Equations?

Factorising is like breaking down a difficult problem into smaller, easier pieces. When we factorise a quadratic equation, we're looking for two simpler expressions that multiply together to give us the original equation. Once we have these, solving for xx becomes straightforward.

The Guide Number Method

The Guide Number Method is a step-by-step way to factorise quadratic equations. It's especially helpful when the number in front of x2x^2 (called the coefficient) is not 1. Let's go through each step with an example.


infoNote

Example Problem:

Solve the quadratic equation: 6x2+11x10=06x^2 + 11x - 10 = 0

We'll solve this equation step by step using the Guide Number Method.


Step 1: Find the Guide Number

What we do:

Multiply the number in front of x2x^2 (here it's 6) by the last number (here it's -10).

Why we do it:

This multiplication gives us a new number, called the Guide Number. We'll use this Guide Number to help break the middle term into two smaller numbers.

Guide Number=6×(10)=:highlight[60]\text{Guide Number} = 6 \times (-10) = :highlight[-60]

So, the Guide Number is -60.


Step 2: List All Factors of the Guide Number

What we do:

We list out all the pairs of numbers that multiply to give us the Guide Number (-60). These pairs are called factors.

Why we do it:

We need to find two numbers that not only multiply to give -60 but also add up to the middle number in the equation, which is 11. Listing all factors helps ensure we don't miss the right pair.

Factors of -60 are: (1,60)(-1, 60) (1,60)(1, -60) (2,30)(-2, 30) (2,30)(2, -30) (3,20)(-3, 20) (3,20)(3, -20) (4,15)(-4, 15) (4,15)(4, -15) (5,12)(-5, 12) (5,12)(5, -12) (6,10)(-6, 10) (6,10)(6, -10)


Step 3: Identify Which Factors Add Up to the Middle Number

What we do:

Now, we look through the list of factors to find the pair that adds up to 11 (the number in front of xx.

Why we do it:

The correct pair will help us split the middle term into two simpler terms, making the equation easier to factorise.

Looking at our factors, the correct pair is: 15×(4)=60and15+(4)=1115 \times (-4) = -60 \quad \text{and} \quad 15 + (-4) = 11

So, the numbers we need are 15 and -4.


Step 4: Rewrite the Middle Term Using These Factors

What we do:

We replace the middle term 11x in the original equation with the two numbers we found (15 and -4). This gives us:

6x2+15x4x10=06x^2 + 15x - 4x - 10 = 0

Why we do it:

By breaking the middle term into two parts, we make it possible to group and factor the equation in the next step.


Step 5: Factor by Grouping

What we do:

We now group the terms into two pairs and factor each pair separately.

(6x2+15x)(4x+10)=0(6x^2 + 15x) - (4x + 10) = 0

For the first group 6x2+15x6x^2 + 15x, we can take out a common factor, which is 3x: 3x(2x+5)3x(2x + 5)

For the second group 4x10-4x - 10, we can take out -2: 2(2x+5)-2(2x + 5)

So, now the equation looks like this: 3x(2x+5)2(2x+5)=03x(2x + 5) - 2(2x + 5) = 0

Finally, because both parts have a common factor (2x + 5), we factor that out: (3x2)(2x+5)=0(3x - 2)(2x + 5) = 0

Why we do it:

Grouping and factoring step by step helps us break the equation into two binomials (simpler expressions). Once we have these binomials, solving for xx is easy.


Step 6: Solve for xx

What we do:

Set each binomial equal to zero and solve for xx:

  1. 3x2=03x - 2 = 0 3x=2x=233x = 2 \quad \Rightarrow \quad x = \frac{2}{3}
  2. 2x+5=02x + 5 = 0 2x=5x=522x = -5 \quad \Rightarrow \quad x = -\frac{5}{2} So, the solutions are: x=:success[23]orx=:success[52]x = :success[\frac{2}{3}] \quad \text{or} \quad x = :success[-\frac{5}{2}]

Why we do it:

Solving each binomial for xx gives us the possible values for x x that make the original equation true. These are the solutions to the quadratic equation.


infoNote

Exam Tip

When solving quadratic equations, remember that you should always expect to find two solutions for xx. This is because the squared term in the equation means that there are usually two values that can make the equation true. Always check that you've found both solutions before finishing the problem.


Summary

Here's a quick recap of what we did:

  1. Find the Guide Number: Multiply the first number by the last number in the equation.
  2. List All Factors: Write down all pairs of numbers that multiply to the Guide Number.
  3. Identify the Correct Factors: Look for the pair that adds up to the middle number.
  4. Rewrite the Middle Term: Split the middle term using the two factors.
  5. Factor by Grouping: Group the terms and factor each group, then factor out the common binomial.
  6. Solve for xx : Set each factor equal to zero and solve.

By following these steps, you can confidently solve quadratic equations using the Guide Number Method. Remember, the key is to take it one step at a time and practice regularly!

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