2.5 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:
6 y 2 y = 7 1 y

Solution:

6y2 y = 7 1y ( 6y2 )( 1y )=7y 6y+6 y 2 2+2y7y=0 6 y 2 11y2=0 ( 6y+1 )( y2 )=0 6y+1=0        or        y=2 y= 1 6

Question 7:
Solve the equation:
4 m 7 = m ( 8 m 9 )

Solution:
4m 7 =m( 8m9 ) 4m=7m( 8m9 ) 4m=56 m 2 63m 56 m 2 63m4m=0 56 m 2 67m=0 m(56m67)=0 m=0        or        56m67=0                                        m= 67 56

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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