Solving Quadratic Equations by Factorisation Slideshow 19 　Mathematics Mr Richard Sasaki　　

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.)

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.

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 .

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=− 2 2 𝑥=− 3 2 or 𝑥= 2 2 𝑥=− 7 2 or 𝑥= 4 6 3 𝑥= 2 7 5 or 𝑥= −5 10 4

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

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

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=− 5 2 or 𝑥=− 4 3 𝑥=0 or 𝑥=− 5 2 𝑥=0 or 𝑥= 1 4 𝑥= 2 3 or 𝑥=−14 𝑥= 3 4 or 𝑥=− 1 2 𝑥=− 1 2 𝑥= 4 3 𝑥=− 3 2

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?

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.

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

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−2 2 𝑥+2 2 =0 𝑥=± 7 𝑥=±2 2 4 𝑥− 2 𝑥+ 2 =0 3 𝑥−3 5 𝑥+3 5 =0 𝑥=± 2 𝑥=±3 5 2 𝑥− 2 2 𝑥+ 2 2 =0 𝑥=± 2 2 24 𝑥− 2 4 𝑥+ 2 4 =0 𝑥=± 2 4

Answers - Bottom 2 𝑥− 6 6 𝑥+ 6 6 =0 3 𝑥− 6 12 𝑥+ 6 12 =0 𝑥=± 6 6 2 𝑥− 6 6 𝑥+ 6 6 =0 3 𝑥− 6 12 𝑥+ 6 12 =0 𝑥=± 6 6 𝑥=± 6 12 5 𝑥− 3 60 𝑥+ 3 60 =0 2 𝑥− 5 4 𝑥+ 5 4 =0 𝑥=± 3 60 𝑥=± 5 4 The solution for 𝑥 would be imaginary.