Page 5 - TOP ONE ADDMATHS TG4

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

4. Form quadratic equations by using the given roots. PL 3
Bentukkan persamaan kuadratik dengan menggunakan punca-punca yang diberikan.

Example (a) 1 and / dan 6 (b) –1 and / dan 1
2 and / dan –5 4
Let a = 1 and b = 6. 1
Let a = 2 and b = –5. Let a = –1 and b = .
4
Sum of roots, a + b Sum of roots, a + b = 1 + 6 1
= 2 + (–5) = 7 Sum of roots, a + b = –1 + 4
= –3 Product of roots, ab = 1 × 6 = – 3
Product of roots, ab = 6 4
= 2 × (–5) 1
= –10 x – (a + b)x + (ab) = 0 Product of roots, ab = –1 × 4
2
2
2
x – (a + b)x + (ab) = 0 x – 7x + 6 = 0 = – 1
4
2
x – (–3)x + (–10) = 0 2
x + 3x – 10 = 0 x – (a + b)x + (ab) = 0
2

1
x – – 3  x – = 0
2
Alternative Method 4 4
3
1
(x – 2)(x + 5) = 0 x + x – = 0
2
x + 5x – 2x – 10 = 0 4 4
2
2
x + 3x – 10 = 0 4x + 3x – 1 = 0
2
5. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.
Example
If a and b are the roots of the quadratic equation 2x – x – 15 = 0, form new quadratic equations using the
2
roots
Jika a dan b ialah punca-punca bagi persamaan kuadratik 2x – x – 15 = 0, bentukkan persamaan kuadratik yang baru
2
menggunakan punca-punca
2
(i) 2 and / dan (ii) (a + 3) and / dan (b + 3)
a b
2
2x – x – 15 = 0 (i) Sum of roots: (ii) Sum of roots:
2
2 + = 2b + 2a (a + 3) + (b + 3) = a + b + 6
b a b ab
a + b = – = 1 + 6
a 2(a + b) 2
= 13
= 1 ab = 2
2 1
2  
c = 2 Product of roots:
ab =
a – 15 (a + 3)(b + 3) = ab + 3a + 3b + 9
= – 15 2 = ab + 3(a + b) + 9
2 = – 2 15 1
15 = – + 3   + 9
2 2
Product of roots: = 3
4
2
2
   = ab x – x + 3 = 0
2 13
b
a
2
4 2x – 13x + 6 = 0
2
=
– 15
2
= – 8
15
2
8


x – – 15  x + – 15  = 0
2
2
x + 2 x – 8 = 0
15 15
15x + 2x – 8 = 0
2
© Penerbitan Pelangi Sdn. Bhd. 20
02 TOP 1 + MATH F4.indd 20 20/12/2019 9:34 AM