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# 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 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 7

Hence, 9 and -9 are the roots of the equation

### 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 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 33

Hence, 1 and are the roots of the given equation

### Solution 37

Hence, are the roots of the given equation

### Solution 38

Hence, 2 and are the roots of given equation

### Solution 46

Hence, are the roots of the given equation

### 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 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 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. 4C

### 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 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

## Quadratic Equations Exercise Ex. 4D

### Solution 8

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

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

### 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 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 38(i)

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

1 year ago

### Solution 41

Let the present age of Meena be x

Then,

Hence the present age of Meena is 7 years

### 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 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 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 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 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