1. Solving Quadratic Equations by Factorisation
Slideshow 　Mathematics
Mr Richard Sasaki　　

2. Objectives
Understand the solutions for a quadratic in the form 𝑥+𝑎 𝑥+𝑏 =0
Be able to factorise quadratics to help solve them
Be able to solve quadratic equations where solutions may be in surd form (no 𝑥 co-efficient.)

3. Factorised Quadratics
Factorising quadratic equations is the quickest way of solving simple cases.
After we factorise when the 𝑥 2 coefficient is 1, we get something in the form 𝑥+𝑎 𝑥+𝑏 =0. What are the solutions?
𝑥=−𝑎 or 𝑥=−𝑏
Why?
If some 𝑚∙𝑛=0, either 𝑚=0 or 𝑛=0, right?
For the same reason, if 𝑥+𝑎 𝑥+𝑏 =0, either 𝑥+𝑎=0 or 𝑥+𝑏=0. So 𝑥=−𝑎 or 𝑥=−𝑏.

4. Other Quadratics What are the solutions for 𝑎 𝑥+𝑏 𝑥+𝑐 =0?
We can divide by 𝑎 and still get 𝑥+𝑏 𝑥+𝑐 =0 so 𝑥=−𝑏 or 𝑥=−𝑐.
What are the solutions for 𝑎 𝑏𝑥+𝑐 𝑚𝑥+𝑛 =0?
We can divide by 𝑎 and get 𝑏𝑥+𝑐 𝑚𝑥+𝑛 =0.
We would then get
𝑏𝑥+𝑐=0 or
𝑚𝑥+𝑛=0
If we make 𝑥 the subject,
𝑥=− 𝑐 𝑏 or
𝑥=− 𝑛 𝑚
Let’s practice. Solve…
𝑥+3 𝑥−4 =0
𝑥=−3 or 𝑥=4.
2 𝑥+7 𝑥−1 =0
𝑥=−7 or 𝑥=1.
3𝑥+2 2𝑥−5 =0
𝑥=− 2 3 or 𝑥= 5 2 .

5. 𝑥=−2 or 𝑥=9 𝑥=3 or 𝑥=6 𝑥=−5 or 𝑥=−4 𝑥=8 𝑥= 1 2 or 𝑥=−3 𝑥=3 or 𝑥=0
𝑥=−1
𝑥= 1 2 or 𝑥=−3
𝑥=3 or 𝑥=0
𝑥=−2 or 𝑥=0
𝑥= 2 3 or 𝑥=− 7 2
𝑥= 9 2 or 𝑥=−5
𝑥= 3
𝑥=− 5 or 𝑥=− 2
𝑥= 3 or 𝑥=−
𝑥=− or 𝑥=
𝑥=− or 𝑥=
𝑥= or 𝑥= −

6. Factorisation
Let’s solve equations in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 where 𝑎, 𝑏, 𝑐∈ℝ.
If 𝑎 𝑥 2 +𝑏𝑥+𝑐=0, we can solve for 𝑥 after factorising.
Example
Factorise and hence, solve 𝑥 2 −5𝑥−14=0.
𝑥+2 𝑥−7 =0
∴𝑥=−2 or 7.
Example
Factorise and hence, solve 2 𝑥 2 +7𝑥−15=0.
𝑎∙𝑏=−30, 𝑎+𝑏=7
⇒𝑎= , 𝑏= 10 −3
2 𝑥 2 +10𝑥−3𝑥−15=0 ⇒
2𝑥 𝑥+5 −3 𝑥+5 =0⇒
(2𝑥−3)(𝑥+5)=0⇒
𝑥= 3 2 , 𝑥=−5

7. Answers - Easy 𝑥=−3 or 𝑥=−1 𝑥=2 or 𝑥=−1 𝑥=−5 or 𝑥=4 𝑥=3 or 𝑥=5 𝑥=±2
𝑥=±7
𝑥=−7
𝑥=4
𝑥=−9 or 𝑥=8
𝑥=10 or 𝑥=−3
𝑥=−13 or 𝑥=8
𝑥=9 or 𝑥=16
𝑥=−5 or 𝑥=0
𝑥=7 or 𝑥=0
𝑥=−7 or 𝑥=−9
𝑥=−5 or 𝑥=3

8. Answers - Hard 𝑥=−5 or 𝑥=3 𝑥=−2 or 𝑥=8 𝑥=−4 or 𝑥=2 𝑥=−7 or 𝑥=3
𝑥=± 1 2
𝑥=± 3 4
𝑥=− or 𝑥=− 4 3
𝑥=0 or 𝑥=− 5 2
𝑥=0 or 𝑥= 1 4
𝑥= or 𝑥=−14
𝑥= or 𝑥=− 1 2
𝑥=− 1 2
𝑥= 4 3
𝑥=− 3 2

9. Factorisation
We know how to solve equations in the form 𝑎𝑥 2 =𝑐. We should consider the root symbol to imply just the positive square root.
Example
Solve 2 𝑥 2 =12.
2 𝑥 2 =12⇒
𝑥 2 =6⇒
𝑥=± 6
How can we write the expression in the form 𝑎 𝑥+𝑏 𝑥+𝑐 =0?

10. No 𝑥− coefficient Example
Write 2 𝑥 2 =12 in the form 𝑎 𝑥−𝑏 𝑥+𝑏 =0 where 𝑎∈ℤ, 𝑏∈ℝ. Do not simplify.
2 𝑥 2 =12⇒
2 𝑥 2 −12=0
⇒2 (𝑥 2 −6)=0
⇒2(𝑥+ 6 )(𝑥− 6 )=0
As there is no 𝑥 coefficient, we can use the principle that 𝑥 2 − 𝑦 2 = 𝑥+𝑦 𝑥−𝑦 .
Note: Never simplify! You should always write 𝑎 as the 𝑥 2 − coefficient. So for the above question the answer is not (𝑥+ 6 )(𝑥− 6 )=0.

11. No 𝑥− coefficient Let’s try one more example. Example
Write 3 𝑥 2 =60 in the form 𝑎 𝑥−𝑏 𝑥+𝑏 =0 and solve where 𝑎∈ℤ, 𝑏∈ℝ. Do not simplify.
3 𝑥 2 =60⇒
3 𝑥 2 −60=0
⇒3 (𝑥 2 −20)=0
⇒3(𝑥+ 20 )(𝑥− 20 )=0
⇒3(𝑥+2 5 )(𝑥−2 5 )=0
⇒𝑥=±2 5

12. Answers - Top 2 𝑥− 7 𝑥+ 7 =0 3 𝑥−2 2 𝑥+2 2 =0 𝑥=± 7 4 𝑥− 2 𝑥+ 2 =0
2 𝑥− 7 𝑥+ 7 =0
3 𝑥− 𝑥 =0
𝑥=± 7
𝑥=±2 2
4 𝑥− 2 𝑥+ 2 =0
3 𝑥− 𝑥 =0
𝑥=± 2
𝑥=±3 5
2 𝑥− 𝑥 =0
𝑥=±
24 𝑥− 𝑥 =0
𝑥=±

13. Answers - Bottom 2 𝑥− 6 6 𝑥+ 6 6 =0 3 𝑥− 6 12 𝑥+ 6 12 =0 𝑥=± 6 6
2 𝑥− 𝑥 =0
3 𝑥− 𝑥 =0
𝑥=±
𝑥=±
5 𝑥− 𝑥 =0
2 𝑥− 𝑥 =0
𝑥=±
𝑥=±
The solution for 𝑥 would be imaginary.