Math For Studens.
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Online Math
Exercise No 1.1 (Solutions Exercise)
10th Class math Solution Online (Science Group)
Matric, FSC & BSC
Available online at: mathforstudents.org
Quadratic Equation
Quadratic Equation:
An equation, which contain the square of the un-known (variable)
quantity but no higher power, is called a quadratic equation or a equation of the second
degree.
Solution By Factorization
ii)
Example 1:
3x2-6x = x + 20
Solution:
3x2-6x-x-20 = 0
3x2 7x 20 = 0
3x (x 4) + 5(x 4) = 0
(x 4) (3x + 5) = 0
x4=0
,
3x + 5 = 0
x=4
,
x = -5/3
Ans
iii)
5
S .S 4,
3
Example: 2
5 x 2 30 x
Solution:
5 x 2 30 x 0
5 x ( x 6) 0
x6 0
S .S 0, 6
x2 4 x
1
3
7
7 x 2 28 3 x
1
21
7 x2 28 3x 21
7 x2 28 21 3x 0
7 x 2 7 3x 0
7 x2 3x 7 0
2
x 1
6
x 1
x
x 2 ( x 1)( x 1)
6
x( x 1)
x 2 ( x 1) 2
6
x2 x
x 2 x 2 1 2( x)(1)
6
x2 x
2 x2 1 2 x
6
x2 x
2 x 2 1 2 x 6( x 2 x)
2 x2 1 2 x 6 x2 6 x
2 x2 1 2 x 6 x2 6 x 0
4x2 1 14x 0
4x2 14x 1 0
Exercise No 1.1
1- Write the following quadratic
equation in the standard form and
point out pure quadratic equation.
i)
(x + 7) (x 3) = -7
(4 x 2 14 x 1) 0
Ans
x 3x 7 x 21 7
x2 4 x 21 7
x2 4x 21 7 0 s
x2 4x 14 0
2
4x2 14x 1 0
MATH FOR STUDENTS
Matric Math Solution Ex 1.1
x4 x2
40
x2
x
( x)( x 4) ( x 2) 4( x 2)( x)
0
( x)( x 2)
12(2x2 8x 7) 25( x2 5x 6)
iv)
x 2 4 x x 2 (2) 2 2( x)(2) 4 x 2 8 x
x2 2x
24 x 2 96 x 84 25 x 2 125 x 150
x 2 29 x 66 0
2)
i)
x 2 x 20 0
x 2 5 x 4 x 20 0
x ( x 5) 4( x 5) 0
( x 5)( x 4) 0
x2 4 x x2 4 4 x 4 x2 8x
0
x2 2x
x 5 0
,
x5
,
S .S 4,5
x 2 4x x 2 4 4 x 4 x 2 8 x 0
x 2 11x 152 0
x 2 19 x 8 x 152 0
x ( x 19) 8( x 19) 0
( x 8)( x 19) 0
8x 8x 4 x 2 4 0
4x2 4 0
x 8 0
,
x 8
,
S .S 8,19
4( x2 1) 0
x2 1 0
x 19 0
x 19
x 2 x 20 0
x 5 x 4 x 20 0
x ( x 5) 4( x 5) 0
( x 4)( x 5) 0
iii)
2
x 2 3 x ( x 2 4 x 5 x 20)
1
x2 4x
x40
,
x 4
,
S .S 4,5
x 2 3x x 2 4 x 5 x 20
1
x2 4x
iv)
x 5 0
x5
3 y 2 y( y 5)
3 y2 y2 5 y
3 x 4 x 5 x 20 x 2 4 x
3 y 2 y 2 5 y
4 x 20 x 4 x 0
2
2 y2 5 y
x 2 4 x 4x 20 0
2 y2 5 y 0
y (2 y 5) 0
2y 5 0
2 y 5
5
y
2
x 2 20 0
Pure quadratic eq.
x40
x 4
ii)
4x 4 4x 4x2 8x 0
Pure quadratic eq.
x3 x5
v)
1
x4
x
x( x 3) ( x 5)( x 4)
1
x( x 4)
Solve by Factorization
x 2 20 0
x 1 x 2 25
x 2 x 3 12
( x 1)( x 3) ( x 2)( x 2) 25
( x 2)( x 3)
12
vi)
5
S .S 0,
2
v)
4 35 x 17 x 2
x 2 3 x x 3 x 2 2 x 2 x 4 25
x 2 3x 2 x 6
12
17 x 2 35 x 4 0
(17 x2 35x 4) 0
x 2 8 x x 2 7 25
x2 5x 6
12
17 x 2 35 x 4 0
17 x 2 32 x 2 x 4 0
17 x ( x 2) 2( x 2) 0
( x 2)(17 x 2) 0
2 x 2 8 x 7 25
x2 5x 6
12
Matric Math Solution Ex 1.1
x20
x 2
MATH FOR STUDENTS
17 x 2 0
2
x
17
2 x 2 10 x 3 x 15 0
Taking common
2 x( x 5) 3( x 5) 0
(2 x 3)( x 5) 0
2x 3 0
x 5 0
,
3
x5
x
,
2
3
S .S ,5
2
3)
Solve the following equation by
completing square.
i)
7 x2 2x 1 0
Solve:
Dividing by 7
7 x2 2 x 1 0
7
7 7 7
2x 1
x2
0
7 7
By completing square
S .S 2,
17
x 1
x
25
vi)
x
x 1 12
( x 1)( x 1) x 2 25
x( x 1)
12
x 2 x x 1 x 2 25
x2 x
12
2 x 2 2 x 1 25
x2 x
12
by cross multiply
12(2 x 2 2 x 1) 25( x 2 x)
24 x 2 24 x 12 25 x 2 25 x
24 x 2 24 x 12 25 x 2 25 x 0
1 1
1 1
( x) 2 2( x)
7 7
7 7
x 2 x 12 0
( x 2 x 12) 0
1
1 1
x
7
7 49
x x 12 0
2
x 2 4 x 3 x 12 0
x( x 4) 3( x 4) 0
1 7 1
x
7
49
Taking square root
2
( x 3)( x 4) 0
x 3 0
x3
,
,
x40
1
8
x
7
49
x 4
1 8
7
7
8 1
x
7
7
8 1
x
7
1 2 2
x
Ans
7
ax 2 4 x a 0,
ii)
S .S 3, 4
2
1
1
vii)
x9 x3 x4
2
1( x 4) 1( x 3)
x9
( x 3)( x 4)
2
x 4 x 3
2
x 9 x 4 x 3x 12
2
1
2
x 9 x 7 x 12
By cross multiply
a0
Do yourself as above
2( x 2 7 x 12) 1( x 9)
2 x 2 14 x 24 x 9
2 x 2 14 x 24 x 9 0
2 x 2 13 x 15 0
Factorization
3
Matric Math Solution Ex 1.1
MATH FOR STUDENTS
11x 2 34 x 3 0
iii)
lx 2 mx n 0,
iv)
Slove:
Dividing by l
lx 2 mx n
0
l
l
l
mx
n
x2
0
l
l
mx
n
x2
l
l
By completing square
Solve:
Dividing by 11
11x 2 34 x 3 0
11
11 11 11
x2
34 x 3
0
11 11
x2
34 x
3
11
11
l0
n m
m m
( x) 2( x)
l 2l
2l 2l
By completing square
m
n m2
2l
l 4l 2
3 34
34 34
( x) 2 2( x)
11 22
22 22
2
m
4nl m 2
2l
4l 2
Taking square root
2
34
3 1156
x
22
11 484
m
4nl m 2
x
2l
4l 2
34
132 1156
x
22
484
34 1024
x
22
484
m
4nl m2
2l
2l
4nl m2 m
2l
2l
4nl m 2 m
2l
Taking square root on both sides
2
34
1024
x
22
484
34
32
22
22
34 32
22 22
32 34
22
66
22
x3
,
,
x
x
m 4nl m 2
x
Ans
2l
v)
3x 2 7 x 0
Solve:
By dividing 3
3 x2 7 x
0
3
3
7x
x2
0
3
By completing square
20
22
7 7
7
( x) 2( x) 0
6 6
6
1
11
7
49
x
6
36
Taking square root
1
S .S 3,
11
7
49
x
6
36
Ans
4
7
7
6
6
MATH FOR STUDENTS
Matric Math Solution Ex 1.1
x0 ,
7
3
3 x 2 12 x 9
3
3
3
7
S .S 0,
3
2
vi)
x 2 x 195 0
by completing square
2
( x) 2( x)(1) (1)2 195 (1)2
Same as above do yourself
15 7
vii)
x2 x
2 2
Solve:
15 7
x x2
2 2
7
15
x2 x
2
2
By completing square
2
7 7 15 7
( x) 2( x)
2 4
4 4
x 2 4 x 3
By completing square
( x)2 2( x)(2) (2)2 3 (2)2
(x 2)2 3 4
(x 2)2 1
Now do yourself as above
7( x 2a)2 3a 2 5a(7 x 23a)
x)
7( x 2 4a 2 4ax) 3a 2 35ax 115a 2
7 x 2 28a 2 28ax 3a 2 35ax 115a 2
7 x 2 28a 2 28ax 3a 2 35ax 115a 2 0
7 x 2 84a 2 7 ax 0
7 x 2 7 ax 84a 2
Dividing by 7
7 x 2 7 ax 84a 2
7
7
7
2
2
x ax 12a
By completing square
7 15 49
x
4
2 16
Now do yourself same as above
33
x 2 17 x
viii)
4
by completing square
33 17
17 17
( x) 2( x)
4 2
2 2
2
a a
a
( x) 2( x) 12a 2
2 2
2
2
a
a2
2
x
12
a
2
4
a
48a 2 a 2
2
4
a
49a 2
2
4
Taking square root
17
33 289
x
2
4
4
Now do yourself Same as above
8
3x 2 5
ix)
4
3x 1 3x 1
Solve:
3x 2 5
8
4
3x 1 3x 1
3x 2 5 8
4
3x 1
3 x 2 13
4
3x 1
Cross multiply
4(3x 1) 3x 2 13
2
a
49a 2
x
2
4
a
7a
x
2
2
7a a
x
2 2
7a a
8a
x
x
4a
2
2
x 3a
S .S 4a, 3a
Ans
3 x 2 13 12 x 4
3 x 2 12 x 4 13
3 x 2 12 x 9
Dividing by 3
The End
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