Permai Plus School

Term2-Test 2 (November,2011)
Additional Mathematics

Name :__________________ Marks :

Class : Sec.2 50
Date :__________________
Parent’s Signature:

________________

1. Find the value of the constant k for which the equation 4 x 2+ k 2+ 1=4 ( k−3 ) x has equal roots.
(3 marks)

Answer:
2 2
4 x + k + 1=4 kx−12 x
2 2
4 x −4 kx+ 12 x +k +1=0
2 2 (1.5m)
4 x + ( 12−4 k ) x +k +1=0
2
∴ a=4 ; b=12−4 k ; c=k +1
2
b −4 ac=0
( 12−4 k )2−4 ( 4 ) ( k 2+1 )=0
(1m)
144−96 k+ 16 k 2−16 ( k 2 +1 )=0
2 2
144−96 k+ 16 k −16 k −16=0
128−96 k =0
1
k =1
3 (0.5m)
1
∴ the value of the constant k is1
3

2. Find the range of values of k for which the equation x 2−2 kx +k 2 −2 k=6 has real roots. (3
marks)

Answer:
2 2
x −2 kx +k −2 k=6
(1.5m)
x 2−2 kx + ( k 2−2 k−6 )=0
2
a=1 ; b=2 k ; c =k −2 k −6

D≥0
(−2 k )2−4 ( 1 ) ( k 2 −2 k−6 ) ≥ 0 (1m)
2 2
4 k −4 k +8 k +24 ≥ 0
8 k ≥−24
k ≥−3
∴ the range value of k ≥−3 (0.5m)

1
3. Given that the equation p x 2+3 px+ p +q=0, where p ≠ 0, has two equal real roots, find q in
terms of p. (3 marks)

Answer:
2
p x +3 px+ p +q=0 (1.5m)
a= p ; b=3 p ; c= p+ q

D=0
( 3 p )2−4 ( p ) ( p+ q ) =0
2 2
9 p −4 p −4 pq=0
2
5 p −4 pq=0 (1m)
p ( 5 p−4 q ) =0
4q
p=0∨ p=
5
5p
q=
4
5p (0.5m)
∴ the q is
4

4. Form a quadratic equation with each set of given roots.

1
a. and −4 (2
3
marks)
Answer:

( ) 1
x− ( x−(−4 ) )=0
3 (1m)
( 3 x−1 ) ( x+ 4 )=0
2
3 x +12 x−x−4=0 (1m)
2
3 x +11 x −4=0

−2 1
b. and (2
9 2
marks)

Answer:

( ( ))( )
x−
−2
9
x−
1
2
=0 (1m)

( 9 x +2 ) ( 2 x−1 ) =0
2 (1m)
18 x −9 x+ 4 x−2=0
2
18 x −5 x−2=0

2
3
5. If the expression ( 1−3 x ) ( x+ 4 )+ p is always negative, find the range of values of the constant
p. (3 marks)

Answer:
( 1−3 x ) ( x+ 4 )+ p=( x +4−3 x 2−12 x ) + p
2
¿−3 x −12 x + x+ 4+ p (1.5m)
2
¿−3 x −11x + 4+ p
a=−3 ; b=−11; c=4+ p

D>0
121−4 (−3 ) ( 4 + p ) >0 (1m)
121+48+ 12 p> 0
169+12 p >0
169>−12 p
−12 p <169
169
p←
12
169 (0.5m)
The range of values of the constant p ←
12

6. Find the discriminant and hence determine the nature of the roots of each of the following
equations
a. ( 2 x−1 ) ( x +1 ) =2 (3 marks)
Answer :

( 2 x−1 ) ( x +1 ) =2
2
2 x +2 x−x−1=2 (1.5m)
2
2 x + x−3=0
a=2 ; b=1 ; c=−3

D=1−4 ( 2 )(−3 )
D=1+ 24 (1m)
D=25
(0.5m)
Since D>0, the equation has two real and distinct roots.

b. x 2−2 x−5=0 (3 marks)

Answer:
2
x −2 x−5=0 (1.5m)
a=1 ; b=−2; c =−5

D=4−4 ( 1 )(−5 )
D=4+ 20 (1m)
D=24
Since D>0, the equation has two real and distinct roots. (0.5m)

4
7. The quadratic equation x 2+ 2 kx+ ( k−1 )( k −3 )=0, has no real roots. Find the range of values of
k. (3 marks)

Answer:
2
x + 2 kx+ ( k−1 )( k −3 )=0
2 2
x + 2 kx+ k −3 k −k +3=0 (1.5m)
2 2
x + 2 kx+ k −4 k +3=0
2 2
x + 2 kx+ k −4 k +3=0
2
a=1 ; b=2 k ; c =k −4 k +3

D<0
( 2 k )2−4 ( 1 ) ( k 2−4 k + 3 ) <0 (1m)
2 2
4 k −4 k +16 k−12<0
16 k −12>0
16 k >12
12
k<
16
3
k<
4 (0.5m)
3
The range of values of k <
4

8. Find the values of c for which the equation 4 x 2−( c+ 2 ) x +c=1 has equal roots. (4 marks)

Answer:
2
4 x −( c+ 2 ) x +c=1
4 x 2−( c+ 2 ) x +c−1=0 (1.5m)
a=4 ; b=−( c+ 2 ) ; c=c−1

D=0
( c +2 )2−4 ( 4 ) ( c−1 )=0
2
c +4 c+ 4−16 c +16=0 (2m)
2
c −12 c +20=0
( c−10 )( c−2 )=0
c=10∨c=2
The value of c are 2 or 10 (0.5m)

9. If the expression m x2−5 x +2 is always positive, find the range of values of the constant m.
(3 marks)

Answer:
2
m x −5 x +2=0
a=m ; b=−5 ; c=2 (1.5m)

D<0
25−4 ( m) ( 2 ) <0
25−8 m< 0 (1m)

5
 25< 8 m
8 m>25
25
m>
8
25 (0.5m)
The range of value constant m ¿
8

10. Determine the number of x-intercepts of the graph of each of the following
a. y=x 2 +8 x +16 (2 marks)
Answer:
2
y=x +8 x +16

D=64−4 ( 1 ) ( 16 ) (2m)
D=64−64
D=0
There are 1 point of intersection with the x-axis.

b. y=−3 x 2−7 x + 4 (2 marks)

Answer:
2
y=−3 x −7 x + 4

D=49−4 (−3 )( 4 ) (2m)

D=49+ 48
D=97
There are 2 point of intersection with the x-axis.

11. If ∝ and β are the roots of x 2+ 9 x−4=0 , find the quadratic equation with the given roots, ∝2
and β 2 (4 marks)

Answer:
2
x + 9 x−4=0
−9 −4 (1m)
Since ( ∝+ β )= =−9 and ( ∝ β ) =
1 1

( ∝2 + β 2) =α 2+ 2 αβ + β 2−2 αβ
2
¿ ( ∝+ β ) −2 αβ
2
¿ (−9 ) −2 (−4 )
¿ 81+8
¿ 89 (2m)

( ∝2 β 2 ) = ( ∝ β ) 2
2
¿ (−4 )
¿ 16

x 2+ ( ∝2+ β 2 ) x+ ( ∝2 β 2 )=0 (0.5m)

2
x + 89 x+16=0

6
The quadratic equation is x 2+ 89 x+16=0 (0.5m)

7
12. The roots of the equation 3 x 2−8 x+ 2=0 are ∝ and β . Find the values of
1 1
a. + (3
∝ β
marks)
Answer:
2
3 x −8 x+ 2=0
8
( ∝+ β )= (1m)
3
2
(∝ β )=
3

()
8 (1.5m)
(1 1
+ =
∝ β )( )
β +α
=
3
αβ (0.5m)
2
=4
3
The valueis 4

∝ β
b. + (3 marks)
β α

8
( ∝+ β )=
3 (1m)
2
(∝ β )=
3

∝ β
+
β α
2 2
α +β
¿
αβ
2 2
∝ +2 ∝ β + β −2 αβ
¿
αβ
2
( ∝+ β ) −2 αβ (1.5m)
=
αβ

¿
64
9
−2 ()
2
3
2
3
64 12
−
9 9
¿
2
3

52 3
¿ ×
9 2
26 2
¿ =8
3 3
(0.5m)

8
 2
The value is 8
3

9
13. Given that the quadratic equation x 2−kx +2 k =3.Find the possible values of k for which the
equation has equal roots. (4 marks)

Answer:
2
x −kx +2 k −3=0
a=1 ; b=−k ; c=2 k−3

D=0
2
k −4 ( 1 ) ( 2 k−3 )=0 (3m)
2
k −8 k +12=0
( k −2 ) ( k−6 )=0
k =2∨6

The possible value of k is 2 or 6.

When k =2; x 2−2 x+ 4−3=0

2
x −2 x+1=0
2
(x−1) =0
x=1
(1m)
When k =6; x 2−6 x +12−3=0
2
x −6 x +9=0
( x−3 )=0
x=3

10