(1) The document is the solutions to a 2 hour quadratic expressions and equations worksheet containing 20 problems labeled (a) through (t). (2) Each problem is set up as a quadratic equation and solved by factoring to find the roots. (3) The solutions are provided in fractional or decimal form in 1,1 format with the two roots separated by a comma.

1. PPR Maths nbk
MODUL 4
SKIM TUISYEN FELDA (STF) SPM ‘ENRICHMENT’.
TOPIC: QUADRATIC EXPRESSIONS AND EQUATIONS.
TIME : 2 HOURS
1. Solve the quadratic equation
3x(x − 1)
(a) =x+6
2
(b) (w – 1)2 – 32 = 0
(c) 2a2 = 3(1 + a) + 2
2
5p + 3p
(d) =4
1+p

2. PPR Maths nbk
2
3t
(e) = 7t – 4
2
11x − 5
(f) x2 – 2 =
4
2
m −6
(g) =m
5
(h) (2x + 1)(x – 2)=7
p+2
(i) p + 2 =
p−3

3. PPR Maths nbk
(j) 3x(2x – 1)+ 8x = 1
2
(k) 3y −2 = y
5
(l) (r –1)(r + 3) = 5(r + 3)
(m) 7p – 2p2 = 2(1 + p)
2
x + 25
(n) 5x =
2

4. PPR Maths nbk
2
p +5
(o) =p
6
− 3(5x − 1)
(p) 4(5x – 1) =
x
7 − 6d
(q) d =
d
2
2m + 5m
(r) =2
m +1
3m(m − 1)
(s) =2
m +1

5. PPR Maths nbk
(t) y2 + 9y – 1 = 3(y – 2)

6. PPR Maths nbk
MODULE 4 - ANSWERS
TOPIC: QUADRATIC EXPRESSIONS AND EQUATIONS.
Answer:
3x(x − 1)
(a) =x+6
2
3x2 –x – 12 = 0 ………..1
(3x+ 4)(x – 3) =0 ……….1
4
x= x= 3 …………….1,1
3
(b) (w – 1)2 – 32 = 0
w2– 2w – 8= 0 …………..1
(w+2)(w–4)=0 …………..1
w=4 w= –2 ……………1,1
(c) 2a2 = 3(1 + a) + 2
2a2– 3a – 5 = 0 …………1
(a+1)(2a – 5) = 0………..1
5
a= –1 x = …………..1,1
2
2
5p + 3p
(d) =4
1+p
5p2– p – 4 = 0 …………1
(5p– 4)(p+1) = 0…….....1
4
p= p = – 1………..1,1
5
2
3t
(e) = 7t – 4
2
3t2+ 14t – 8 = 0 ………..1
(3t – 2)(t – 4)=0 ………..1

7. PPR Maths nbk
2
t= t = 4 …………1,1
3
11x − 5
(g) x2 – 2 =
4
2
4x – 11x – 5 = 0 ………1
(4x+1)(x - 3) = 0 ………1
1
x= x=3
4
2
m −6
(g) =m
5
m2– 5m – 6 = 0 ………..1
(m - 6)(m + 1) = 0………1
m=6 m= -1 …………1,1
(i) (2x + 1)(x – 2)=7
2x2– 3x – 9 = 0 …………1
(x+3)(x - 3) = 0 …………1
x = -3 x = 3 ……………. 1,1
p+2
(i) p + 2 =
p−3
p2 – 2p – 8 = 0 …………1
(p+2)(p– 4)=0 ………….1
p= –2 p = 4……………1,1
(k) 3x(2x – 1)+ 8x = 1
6x2+ 5x – 1 = 0 …………1
(x+1)(6x - 1) = 0 ………..1
1
x = -1 x = …………..1,1
6

8. PPR Maths nbk
2
(k) 3y −2 = y
5
2
3y – 5y – 2 = 0………………1
(y+1)(y– 2)=0 ………………..1
y= -1 y = 2 …………………1,1
(o) (r –1)(r + 3) = 5(r + 3)
r2 – 3yr– 18 = 0 ……………..1
(r+6)(r– 3)=0 ………………..1
y= -6 y = 3 ……………….1,1
(p) 7p – 2p2 = 2(1 + p)
2p2– 5p + 2 = 0 …………..1
(2p– 1)(2p+5) = 0 ………..1
1 −5
p= p= …………1,1
2 2
x 2 + 25
(q) 5x =
2
x2 – 10x + 25 = 0 …………1
(x-5)(x-5)=0……………….1
x=5 ………………………..1
2
p +5
(o) =p
6
p2– 6p + 5 = 0 ………….1
(p– 1)(p+5) = 0 …………..1
p=1 p = -5……………..1,1
− 3(5x − 1)
(q) 4(5x – 1) =
x
2
20x – 11x -3 = 0 ……….1
(5x+ 1)(4x-3) = 0 ………..1
1 1
x=- x= ………….1,1
5 4

9. PPR Maths nbk
7 − 6d
(q) d =
d
d2+ 6d -7 = 0 ………..1
(d– 1)(d+7) = 0 ………1
x=1 x = -7 ……….1,1
2
2m + 5m
(r) =2
m +1
2m2+ 3m -2 = 0 ……….1
(2m– 1)(m+1) = 0………1
1
x = -2 x= ………….1,1
2
3m(m − 1)
(s) =2
m +1
3m2- 5m -2 = 0 …………..1
(3m+1)(m-2) = 0 ………….1
1
x=2 x=- …………1,1
3
(u) y2 + 9y – 1 = 3(y – 2)
y2+ 6y+5 = 0 ……………1
(y+1)(y+5) = 0 ………….1
x = -1 x = -5 ………….1,1