PDF file of Assignment 2 complete solution

1. University of Central Punjab
Assignment 2
Q1: Solve the following quadratic equation by completing square method
( ) 2
6 5 6 0
i x x
− − =
( ) 2
16 8 3 0
ii x x
− − =
( ) 2
6 5 6 0
i x x
− − =
( ) ( )
2
2
2
2 2
2
2
2
2
2
Sol:-
6 5 6 0
6 5 6
by dividing 6 on both sides
5
1
6
5 5 5
2 1
12 12 12
5 25
1
12 144
5 144 25
12 144
5 169
12 144
5 169
12 144
5 13
12 12
5 13
12 12
he
x x
x x
x x
x x
x
x
x
x
x
x
− − =
− =
− =
     
− + = +
     
     
 
− = +
 
 
+
 
− =
 
 
 
− =
 
 
 
− = 
 
 
− = 
= 
re
5 13 5 13
12 12 12 12
18 8
5 12
3 2
3
5 3
x OR x
x x
x x
= + = −
−
= =
−
= =
( ) 2
16 8 3 0
ii x x
− − =
Sol:-
( ) ( )
2
2
2
2
2 2
2
2
2
2
2
2
16 8 3 0
16 8 3
by dividing 16 on both sides
8 3
6 6
4 1
3 2
2 2 1 2
2
3 3 2 3
2 1 4
3 2 9
2 1 9 4 2
3 2 9 9 2
2 9 8
3 18 18
2 17
3 18
2 17
3 18
2
3
x x
x x
x x
x x
x x
x
x
x
x
x
x
− − =
− =
− =
− =
     
− + = +
     
     
 
− = +
 
 
 
 
− = +
 
 
 
 
− = +
 
 
 
− =
 
 
 
− = 
 
 
−
17
18
2 17
3 18
x
= 
= 

2. Q2: Reducible to quadratic equation
( )
2 1
3 3
13 36 0
i x x
+ + = , ( )
1 1
2 4
4 5 0
ii x x
− − = , ( ) 3 5 1 2
iii x x
− + − =
2 1
3 3
1
3
2
1
2
3
2
2
3
2
13 36 0
let:
So, the equation will become
13 36 0
Now It becomes a quaderatic equation
x x
x y
x y
x y
y y
+ + =
=
 
=
 
 
=
+ + =
1 1
2 4
1
4
2
1
2
4
1
2
2
2
4 5 0
Let:
So, the equation will become
4 5 0
Now it becomes a quaderatic equation
x x
x y
x y
x y
y y
− − =
=
 
=
 
 
=
− − =
( )
( ) ( ) ( )( )
( )( )
( ) ( )
( ) ( ) ( )( )
2
2
2 2
2
2
2
2
2
2
2 2
2
2 2
2 2
2
3 5 1 2
:
3 5 1 2
3 5 1 2 3 5 1 4
3 5 1 2 3 5 1 4
4 6 2 3 3 5 5 4
3 8 5 4 6 4
3 8 5 4 2
3 8 5 4 2
3 8 5 4 2 2 4 2
3 8 5 16 4 16
16 3 4 5 16 8 0
13 8
x x
Sol
x x
x x x x
x x x x
x x x x
x x x
x x x
x x x
x x x x
x x x x
x x x x
x x
− + − =
− + − =
− + − + − − =
− + − + − − =
− + − − + =
− + = − + −
− + = − +
− + = − +
− + = − + + −
− + = + −
− + − − + =
− 1 0
Now it becomes a quaderatic equation
− =
( ) ( )
2
Q3: Let , be the roots of 4 5 6 0, Find equation whose roots are
, 3, 3
3 3
x x
i ii
 
 
 
− − =
+ +
Sol:
( )
( )
2
From equation
4, 5, 6
1
Sum of the roots ( )
3 3 3
5
1 1
3 3 4
5
12
Product of the roots
3 3 9
1 1 6
9 9 4
6 1
36 6
The standard form of equation is
a b c
b
a
c
a
x x
 
 
  
  
= = − = −
= + = +
−
 
 
= − = −
 
 
   
=
  
= =
  
  
−
   
= =
   
   
− −
= 
− + +
2
2
0
So, the required equation is
5 1
0
12 6
5 1
0
12 6
x x
x
x
=
−
   
− + =
   
   
− − =
Sol:
( )
( )( )
( )
( )
( )
From equation
4, 5, 6
Sum of the roots 3 3 6
5
6 6
4
5 6 4 5 24 29
4 4 4 4
Product of the roots = 3 3
3 3 9 3 9
5
6
3 9 3 9
4 4
5
6 6 5
3 9 3
4 4 4
a b c
b
a
c b
a a
   
 
     
= = − = −
= + + + = + +
− −
−
= + = +
 +
= + = =
+ +
= + + + = + + +
− −
 
− −
   
= + + = + +
 
   
     
− −
 
− −
= + + = +
 
 
( )
2
2
2
9
4
6 15 9 4 6 15 36 45
4 4 4 4 4
The standard form of equation is
0
So, the required equation is
29 45
0
4 4
4 29 45 0
x x
x x
x x
  
 
+
 
 
−  − + +
= + + = =
− + + =
− + =
− + =

3. ( ) ( )
13 1 1 11
Q4: Find if 4, 3 ? 3, 33, 10
i a a d ii d if a a n
= = = = = =
Sol:
( )
( )
( )
13 1
1
13
13
Given that
?, 13 4, 3
: 1
4 13 1 3
4 12 3
4 36
40
n
a n a d
Formula a a n d
by putting values
a
a
= = = =
= + −
= + −
= +
= +
=
Sol:
( )
( )
1 11
1
11 1
Given That
? 3, 33, 10
: 1
11 1
33 3 10
33 3 10
30 10
30
10
3
n
d if a a n
Formula a a n d
a a d
d
d
d
d
d
= = = =
= + −
= + −
= +
− =
=
=
=
( ) ( )
( ) ( )
( ) ( )
( ) ( )
20 3 8
3 1 8 1
1 1
1 1
1 1
Q5: Find 20 term of A.P if 3 term is 7 and 8 term is 17.
:
: ?, 7, 17
3 1 8 1
7 2 17 7
Subtract eq from eq
7 17 2 7
10 2 7
10
th rd th
Sol
Giventhat a a a
a a d a a d
a d i a d ii
ii i
a d a d
a d a d
− = = =
= + − = + −
= + − − − − − − = + − − − − − −
− = + − +
− = + − −
−
( )
( )
( )
( )
1
1
1
20 1
20
5
2
put the value of in eq
7 2 2
7 4
7 4 3
,
20 1
3 19 2
3 38
41
d
d
d i
a
a
a
Now
a a d
a
= −
=
= +
= +
= − =
= + −
= +
= +
=