MODULE 4- Quadratic Expression and Equations
- 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