# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

## Chapter 10 - Quadratic Equations Ex. 4A

_{}

Hence, _{}are the roots of _{}

(i)_{}is a quadratic polynomial

_{}= 0 is a quadratic equation

_{}

Clearly _{} is a quadratic polynomial

_{}is a quadratic equation.

_{} is a quadratic polynomial

_{ }= 0 is a quadratic equation

Clearly, is a quadratic equation

_{}is a quadratic equation

_{}is not a quadratic polynomial since it contains_{} in which power _{} of x is not an integer.

_{}is not a quadratic equation

And Being a polynomial of degree 2, it is a quadratic polynomial.

Hence, _{}is a quadratic equation.

And _{}being a polynomial of degree 3, it is not a quadratic polynomial

Hence, is not a quadratic equation

_{}is not a quadratic equation

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

_{}

Since x = 1 is a solution of _{}it must satisfy the equation.

_{}

Hence the required value of k = -4

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

_{}

_{}

Hence, 9 and -9 are the roots of the equation _{}

_{}

Hence, _{}are the roots of _{}

_{}

Hence, _{}are the roots of equation_{}

_{}

Hence, _{}and 1 are the roots of the equation _{}.

_{}

_{}are the roots of the equation _{}

_{}

Hence, _{}are the roots of the given equation _{}

_{}

Hence, _{}are the roots of given equation

_{}

Hence,_{ }are the roots of the given equation

_{}

Hence, _{}are the roots of the given equation

_{}

Hence, _{}are the roots of the given equation

_{}

Hence, _{}are the roots of the given equation

_{}

Hence, 1 and _{}are the roots of the given equation

_{}

Hence, _{}are the roots of the given equation

_{}

Hence, 2 and _{}are the roots of given equation

_{}

Hence, _{}are the roots of the given equation

_{}

Hence, _{}are the roots of given equation

_{}

Hence, _{}are the roots of given equation

_{}

Hence, _{}are the roots of given equation

Putting _{}the given equation become

_{}

Case I:

_{}

Case II:

_{}

Hence, _{}are the roots of the given equation

The given equation

_{}

Hence, _{}is the roots of the given equation

_{}

Hence, -2,0 are the roots of given equation

_{}

Hence, _{}are the roots of given equation

_{}

Hence, 3 and 2 are roots of the given equation

## Chapter 10 - Quadratic Equations Ex. 4B

## Chapter 10 - Quadratic Equations Ex. 10E

Let the required number be x and x - 3, then

_{}

Hence, the required numbers are (24,21) or (-21 and-24)

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

Let the required number be x and y, hen

_{}

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

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

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

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

Let the numerator and denominator be x, x + 3

Then,

_{}

Hence, numerator and denominator are 2 and 5 respectively and fraction is _{}

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

Let the number of students be x, then

_{}

_{}

Hence the number of students is 50

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)

Let the age of son be x and age of man = y

1 year ago

_{}

Let the present age of Meena be x

Then,

Hence the present age of Meena is 7 years

** **

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

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

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

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

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 = _{}

_{}

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

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

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

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

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

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

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

Let the length = x meter

Area = length breadth = _{}

_{}

If ength of the rectangle = 15 m

_{}

Also, if length of rectangle = 24 m

_{}

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

_{}

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

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

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

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 = _{}

Let the shorter side of triangle be x meter

Then, its hypotenuse = (2x - 1)meter

And let the altitude = (x + 1) meter

_{}

## Chapter 10 - Quadratic Equations Ex. 10F

## Chapter 4 - Quadratic Equations MCQ

## Chapter 4 - Quadratic Equations Ex. 4C

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.

_{}

The given equation is _{}

_{}

For real and equal roots, we must have D = 0

_{}

The given equation is _{}

_{}

For real and equal roots , we must have D =0

_{}

## Chapter 4 - Quadratic Equations Ex. 4D

Let the required number be x and x - 3, then

_{}

Hence, the required numbers are (24,21) or (-21 and-24)

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

Let the required number be x and y, hen

_{}

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

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

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

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

Let the numerator and denominator be x, x + 3

Then,

_{}

Hence, numerator and denominator are 2 and 5 respectively and fraction is _{}

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

Let the number of students be x, then

_{}

_{}

Hence the number of students is 50

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)

Let the age of son be x and age of man = y

1 year ago

_{}

Let the present age of Meena be x

Then,

Hence the present age of Meena is 7 years

** **

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

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

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

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

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 = _{}

_{}

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

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

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

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

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

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

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

Let the length = x meter

Area = length breadth = _{}

_{}

If ength of the rectangle = 15 m

_{}

Also, if length of rectangle = 24 m

_{}

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

_{}

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

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

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

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 = _{}

Let the shorter side of triangle be x meter

Then, its hypotenuse = (2x - 1)meter

And let the altitude = (x + 1) meter

_{}

## Chapter 4 - Quadratic Equations 29

### Other Chapters for CBSE Class 10 Maths

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 16- Areas of Circle, Sector and Segment Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- Probability### R S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects

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