Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 4 - Quadratic Equations
Quadratic Equations Exercise Ex. 4A
Solution 19
Hence, 
are the roots of 
Solution 1(i)
(i)
is a quadratic polynomial
= 0 is a quadratic equation
Solution 1(ii)
Clearly 
is a quadratic polynomial
is a quadratic equation.
Solution 1(iii)
is a quadratic polynomial
= 0 is a quadratic equation
Solution 1(iv)
Clearly, 
is a quadratic equation
is a quadratic equation
Solution 1(v)
is not a quadratic polynomial since it contains
in which power 
of x is not an integer.
is not a quadratic equation
Solution 1(vi)
And 
Being a polynomial of degree 2, it is a quadratic polynomial.
Hence, 
is a quadratic equation.
Solution 1(vii)
And 
being a polynomial of degree 3, it is not a quadratic polynomial
Hence, 
is not a quadratic equation
Solution 1(viii)
is not a quadratic equation
Solution 1(ix)
Solution 1(x)
Solution 1(xi)
Solution 2
The given equation is 
(i)On substituting x = -1 in the equation, we get
(ii)On substituting 
in the equation, we get
(iii)On substituting 
in the equation , we get
Solution 3(i)
Since x = 1 is a solution of 
it must satisfy the equation.
Hence the required value of k = -4
Solution 3(ii)
Since 
is a root of 
, we have
Again x = -2 being a root of , we have
Multiplying (2) by 4 adding the result from (1), we get
11a = 44 
a = 4
Putting a = 4 in (1), we get
Solution 5
Solution 6
Solution 7
Hence, 9 and -9 are the roots of the equation 
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Hence, 
are the roots of 
Solution 14
Hence, 
are the roots of equation
Solution 15
Hence, 
and 1 are the roots of the equation 
.
Solution 16
are the roots of the equation 
Solution 17
Hence, 
are the roots of the given equation 
Solution 18
Hence, 
are the roots of given equation
Solution 20
Solution 21
Solution 22
Hence, 
are the roots of the given equation
Solution 23
Hence, 
are the roots of the given equation
Solution 24
Hence, 
are the roots of the given equation
Solution 25
Hence, 
are the roots of the given equation
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Hence, 1 and 
are the roots of the given equation
Solution 34
Solution 35
Solution 36
Solution 37
Hence, 
are the roots of the given equation
Solution 38
Hence, 2 and 
are the roots of given equation
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Hence, 
are the roots of the given equation
Solution 47
Solution 48
Hence, 
are the roots of given equation
Solution 49
Hence, 
are the roots of given equation
Solution 50
Hence, 
are the roots of given equation
Solution 51
Solution 52
Solution 53
Solution 54
Solution 55(i)
Solution 56
Solution 57
Solution 58
Solution 59(i)
Solution 60
Solution 61
Solution 62
Solution 63
Solution 64(i)
Solution 65
Solution 66
Solution 67
Solution 68
Putting 
the given equation become
Case I:
Case II:
Hence, 
are the roots of the given equation
Solution 69
The given equation
Hence, 
is the roots of the given equation
Solution 70
Solution 71
Hence, -2,0 are the roots of given equation
Solution 72
Hence, 
are the roots of given equation
Solution 73
Hence, 3 and 2 are roots of the given equation
Quadratic Equations Exercise Ex. 4B
Solution 4
Solution 5
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 2
Solution 3
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Quadratic Equations Exercise Ex. 4C
Solution 1(i)
Solution 1(ii)
Solution 1(iii)
Solution 1(iv)
Solution 1(v)
Solution 1(vi)
Solution 2
Solution 3
The given equation is 
This is the form of 
Now 
.
So, the roots of the given equation are real for all real value of p and q.
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8(i)
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
The given equation is 
For real and equal roots, we must have D = 0
Solution 15
The given equation is 
For real and equal roots , we must have D =0
Solution 16
Solution 17
Solution 18
Solution 19(i)
Solution 19(ii)
Solution 19(iii)
Solution 19(iv)
Solution 20
Solution 21
Solution 23
Quadratic Equations Exercise Ex. 4D
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Let the required number be x and x - 3, then
Hence, the required numbers are (24,21) or (-21 and-24)
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Let the smaller part and larger part be x, 16 - x
Then,
-42 is not a positive part
Hence, the larger and smaller parts are 10, 6 respectively
Solution 21
Let the required number be x and y, hen
Solution 22
Let x, y be the two natural numbers and x > y
------(1)
Also, square of smaller number = 4 larger number
---------(2)
Putting value of 
from (1), we get
Thus, the two required numbers are 9 and 6
Solution 23
Let the three consecutive numbers be x, x + 1, x + 2
Sum of square of first and product of the other two
Required numbers are 4, 5 and 6
Solution 24
Let the tens digit be x and units digit be y
Hence, the tens digit is 3 and units digit is (2 3) is
Hence the required number is 36
Solution 25
Let the tens digit and units digits of the required number be x and y respectively
The ten digit is 2 and unit digit is 7
Hence, the required number is 27
Solution 26
Let the numerator and denominator be x, x + 3
Then,
Hence, numerator and denominator are 2 and 5 respectively and fraction is 
Solution 27
Solution 28
Solution 29
Let there be x rows and number of student in each row be x
Then, total number of students = 
Hence total number of student
Total number of students is 600
Solution 30
Let the number of students be x, then
Hence the number of students is 50
Solution 31
Let the marks obtained by Kamal in Mathematics and English be x and y
The marks obtained by Kamal in Mathematics and English respectively are (21,19) or (12,28)
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38(i)
Let the age of son be x and age of man = y
1 year ago
Solution 39
Solution 40
Solution 41
Let the present age of Meena be x
Then,
Hence the present age of Meena is 7 years
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Let the original speed of the train be x km.hour
Then speed increases by 15 km/ph = (x + 15)km/hours
Then time taken at original speed = 
Then, time taken at in increased speed = 
Difference between the two lines taken 
Then, original speed of the train = 45km / h
Solution 49
Solution 50
Let the speed of the Deccan Queen = x kmph
The, speed of other train = (x - 20)kmph
Then, time taken by Deccan Queen = 
Time taken by other train = 
Difference of time taken by two trains is 
Hence, speed of Deccan Queen = 80km/h
Solution 51
Let the speed of stream be x km/h
Speed of boat in still stream = 18 km/h
Speed of boat up the stream = 18 - x km/h
Time taken by boat to go up the stream 24 km = 
Time taken by boat to go down the stream = 
Time taken by the boat to go up the stream is 1 hour more that the time taken down the stream
Speed of the stream = 6 km/h
Solution 52
Let the speed of the stream be x kmph
Then the speed of boat down stream = (8 + x) kmph
And the speed of boat upstream = (8 - x)kmph
Time taken to cover 15 km upstream = 
Time taken to cover 22 km downstream = 
Total time taken = 5 hours
Hence, the speed of stream is 3 kmph
Solution 53
Let the speed of the stream be = x km/h
Speed of boat in still waters = 9 km/h
Speed of boat down stream = 9 + x
time taken by boat to go 15 km downstream = 
Speed of boat upstream = 9 - x
time taken by boat to go 15 km of stream = 
Solution 54
Solution 55
Let the faster pipe takes x minutes to fill the cistern
Then, the other pipe takes (x + 3) minute
The faster pipe takes 5 minutes to fill the cistern
Then, the other pipe takes (5 + 3) minutes = 8 minutes
Solution 56
Solution 57
Solution 58
Let the breadth of a rectangle = x cm
Then, length of the rectangle = 2x cm
Thus, breadth of rectangle = 12 cm
And length of rectangle = (2 12) = 24 cm
Solution 59
Let the breadth of a rectangle = x meter
Then, length of rectangle = 3x meter
Thus, breadth of rectangle = 7 m
And length of rectangle = (3 7)m = 21 m
Solution 60
Let the breadth of hall = x meters
Then, length of the hall = (x + 3) meters
Area = length breadth = 
Thus, the breadth of hall is 14 cm
And length of the hall is (14 + 3) = 17 cm
Solution 61
Solution 62
Let the width of the path be x meters,
Then,
Area of path = 16 10 - (16 - 2x) (10 - 2x) = 120
Hence the required width is 3 meter as x cannot be 10m
Solution 63
Let x and y be the lengths of the two square fields.
4x - 4y = 64
x - y = 16------(2)
From (2),
x = y + 16,
Putting value of x in (1)
Sides of two squares are 24m and 8m respectively
Solution 64
Let the side of square be x cm
Then, length of the rectangle = 3x cm
Breadth of the rectangle = (x - 4) cm
Area of rectangle = Area of square x
Thus, side of the square = 6 cm
And length of the rectangle = (3 6) = 18 cm
Then, breadth of the rectangle = (6 - 4) cm = 2 cm
Solution 65
Let the length = x meter
Area = length breadth = 
If ength of the rectangle = 15 m
Also, if length of rectangle = 24 m
Solution 66
Let the altitude of triangle be x cm
Then, base of triangle is (x + 10) cm
Hence, altitude of triangle is 30 cm and base of triangle 40 cm
Solution 67
Let the altitude of triangle be x meter
Hence, base = 3x meter
Hence, altitude of triangle is 8 cm
And base of triangle = 3x = (3 8) cm = 24 cm
Solution 68
Let the base of triangle be x meter
Then, altitude of triangle = (x + 7) meter
Thus, the base of the triangle = 15 m
And the altitude of triangle = (15 + 7) = 22 m
Solution 69
Let the other sides of triangle be x and (x -4) meters
By Pythagoras theorem, we have
Thus, height of triangle be = 16 cm
And the base of the triangle = (16 - 4) = 12 cm
Solution 70
Let the base of the triangle be x
Then, hypotenuse = (x + 2) cm
Thus, base of triangle = 15 cm
Then, hypotenuse of triangle = (15 +2 )= 17 cm
And altitude of triangle = 
Solution 71
Let the shorter side of triangle be x meter
Then, its hypotenuse = (2x - 1)meter
And let the altitude = (x + 1) meter
Quadratic Equations Exercise MCQ
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 14
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
Solution 53
Solution 54
Solution 55
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