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1.6.2 Quadratic Equations, SPM Paper 2 (Long Questions)


Question 5:
Solve the equation:
(m + 2)(m – 4) = 7(m – 4).
 
Solution:
(m + 2)(m – 4) = 7(m – 4)
m2– 4m + 2m – 8 = 7m – 28
m2– 9m + 20 = 0
(m – 5)(m – 4) = 0
= 5 or m = 4   

Question 6:
Solve the equation:
6y2y=71y

Solution:

6y2y=71y(6y2)(1y)=7y6y+6y22+2y7y=06y211y2=0(6y+1)(y2)=06y+1=0ory=2y=16

Question 7:
Solve the equation:
4m7=m(8m9)

Solution:
4m7=m(8m9)4m=7m(8m9)4m=56m263m56m263m4m=056m267m=0m(56m67)=0m=0or56m67=0   m=6756

Question 8:
Diagram above shows a rectangle ABCD.
(a) Express the area of ABCD in terms of n.
(b) Given the area of ABCD is 60 cm2, find the length of AB.
 
Solution:
(a)
Area of ABCD
= (n + 7) × n
= (n2+ 7n) cm2

(b)
Given the area of ABCD = 60
n2+ 7n = 60
n2+ 7n – 60 = 0
(n – 5) (n + 12) = 0
= 5 or    n = – 12 (not accepted)
 
When n = 5,
Length of AB = 5 + 7 = 12 cm

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