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

Quadratic Equations Exercise Ex. 4A

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)

1 third x squared plus 1 fifth x minus 2 equals 0
rightwards double arrow 5 x squared plus 3 x minus 30 equals 0
Clearly comma space 5 x squared plus 3 x minus 30 equals 0 space is space straight a space quadratic space polynomial.
therefore 1 third x squared plus 1 fifth x minus 2 equals 0 space is space straight a space quadratic space 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)

x plus 2 over x equals x squared
rightwards double arrow x squared plus 2 equals x cubed
rightwards double arrow x cubed minus x squared minus 2 equals 0
And comma space open parentheses x cubed minus x squared minus 2 equals 0 close parentheses space being space straight a space polynomial space of space degree space 3 comma space it space is space not space straight a space quadratic space polynomial.
Hence comma space x plus 2 over x equals x squared space is space not space straight a space quadratic space equation. space

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 root of the equation  , it must satisfy the equation.

Substituting  in  ,

Hence k=-4 and other root is 3.

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 4

Given equation is  .

Substituting   in LHS of the given equation,

  is a solution of the quadratic equation  .

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 19

Hence, are the roots of

Solution 20

Solution 21

Solution 22

rightwards double arrow x equals fraction numerator negative 2 over denominator square root of 3 end fraction space o r space x equals negative 3 square root of 3

Solution 23

Hence, are the roots of the given equation

Solution 24

square root of 7 x squared minus 6 x minus 13 square root of 7 equals 0
rightwards double arrow square root of 7 x squared minus 13 x plus 7 x minus 13 square root of 7 equals 0
rightwards double arrow x open parentheses square root of 7 x minus 13 close parentheses plus square root of 7 open parentheses square root of 7 x minus 13 close parentheses equals 0
rightwards double arrow open parentheses square root of 7 x minus 13 close parentheses open parentheses x plus square root of 7 close parentheses equals 0
rightwards double arrow open parentheses square root of 7 x minus 13 close parentheses equals 0 space or space open parentheses x plus square root of 7 close parentheses equals 0
rightwards double arrow x equals fraction numerator 13 over denominator square root of 7 end fraction space or space x equals negative square root of 7

Solution 25

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 55(ii)

Solution 56

Solution 57

Solution 58

Solution 59(i)

Solution 59(ii)

Solution 60

Solution 61

Solution 62

Solution 63

Solution 64(i)

Solution 64(ii)

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 1(i)

Solution 1(ii)

  

Solution 1(iii)

  

Solution 1(iv)

  

Solution 1(v)

 

Solution 1(vi)

  

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 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 values of p and q.

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8(i)

Solution 8(ii)

The given equation is   

This is of the form  , where

For real and equal roots, we must have

Discriminant, D = 0

       

Solution 9

Solution 10

Solution 11

Since -5 is a root of the quadratic equation  , it must satisfy the equation.

The other quadratic equation is   

  (i)

Substituting  p = 7 in (i),

This is of the form  , where

For equal roots, we must have

Discriminant, D = 0

      

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

The given equation is   

This is of the form  , where

For equal roots, Discriminant, D = 0

Solution 22

The given equation is

This is of the form  , where

Now, Discriminant

Now,

Hence, the given equation has no real roots.

Solution 23

Solution 24

Since   is a root of the quadratic equation  , it must satisfy the equation.

Solution 25

Since   is a root of the quadratic equation  , it must satisfy the equation.

Solution 26

Given quadratic equation is  .

But, square of a number cannot be negative.

Hence, the given equation has no solution.

Solution 27

The given equation is   

This is of the form  , where

For no real roots,

Discriminant, D <0

Solution 28

The given equation is   

This is of the form  , where

Let the roots of the equation be

Then,

Solution 30

The given equation is   

This is of the form  , where

For real and equal roots, we must have

Discriminant, D = 0

Solution 31

The given equation is

This is of the form  , where

For real and equal roots, we must have

Discriminant, D = 0

Solution 32

The given equation is   

This is of the form  , where

For real and equal roots, we must have

Discriminant, D = 0

       

Solution 29

The given equation is

This is of the form  , where

Let the roots of the equation be

Then,

Quadratic Equations Exercise Ex. 4D

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Let the required natural number be x and x - 3.

Then,

x open parentheses x minus 3 close parentheses equals 504
rightwards double arrow x squared minus 3 x minus 504 equals 0
rightwards double arrow x squared minus 24 x plus 21 x minus 504 equals 0
rightwards double arrow x open parentheses x minus 24 close parentheses plus 21 open parentheses x minus 24 close parentheses equals 0
rightwards double arrow open parentheses x minus 24 close parentheses open parentheses x plus 21 close parentheses equals 0
rightwards double arrow x minus 24 equals 0 space or space x plus 21 equals 0
rightwards double arrow x equals 24 space or space x equals negative 21

But, x is a natural number.

rightwards double arrow x not equal to negative 21

Hence, x equals 24 rightwards double arrow x minus 3 equals 24 minus 3 equals 21

Therefore, the required numbers are 24 and 21.

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 part is 10 and the smaller part is 6.

Solution 21

Let the required natural number be x and y.

Then

Therefore, the required natural numbers are 30 and 10 respectively.

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

rightwards double arrow y equals 2 cross times 3 equals 6

Hence, the tens digit is 3 and units digit is 6.

Hence, the required number is 36.

Solution 25

Let the tens and units digits of the required number be x and y respectively.

Then,

x y equals 14
rightwards double arrow y equals 14 over x space space space space space space space space space space.... left parenthesis straight i right parenthesis

Now, the number is 10 x plus y.

So, left parenthesis 10 x plus y right parenthesis plus 45 equals 10 y plus x

rightwards double arrow 9 x minus 9 y equals negative 45
rightwards double arrow x minus y equals negative 5
rightwards double arrow x minus 14 over x equals negative 5 space space space space space space space space space space.... left square bracket From space left parenthesis straight i right parenthesis right square bracket
rightwards double arrow x squared minus 14 equals negative 5 x
rightwards double arrow x squared plus 5 x minus 14 equals 0
rightwards double arrow x squared plus 7 x minus 2 x minus 14 equals 0
rightwards double arrow x open parentheses x plus 7 close parentheses minus 2 open parentheses x plus 7 close parentheses equals 0
rightwards double arrow open parentheses x plus 7 close parentheses open parentheses x minus 2 close parentheses equals 0
rightwards double arrow x plus 7 equals 0 space space or space x minus 2 equals 0
rightwards double arrow x equals negative 7 space or space x equals 2

But, digit cannot be negative.

rightwards double arrow x equals 2
rightwards double arrow y equals 14 over 2 equals 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 students in each row be x.

Then, total number of students =

Hence, total number of students

Therefore, 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 respectively.

Therefore, 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 years and that of man be y years.

1 year ago,

Solution 38(ii)

Let the ages of son and father be x years and y years respectively.

Then,   

Now, sum of the squares of their ages = 1325

But, age cannot be negative.

Therefore, the age of father is 35 years and that of son is 10 years.

Solution 39

Solution 40

Solution 41

Let the present age of Tanvi be x years.

Then,

Hence, the present age of Tanvi 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   km/hr.

Then, increased speed = (x+5)km/hr

Time taken at original speed =   hours

Time taken at increased speed =   hours

Hence, the original speed of the train is 25 km/hr.

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 km/h

Speed of boat in still water = 15 km/hr

Speed of boat upstream = (15-x) km/hr 

Time taken by the boat to go 30 km upstream   hours

Speed of boat downstream =   km/hr

Time taken by the boat to go 30 km downstream

Total time taken by the boat

Hence, the speed of the stream is 5 km/hr.

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 one tap takes x minutes to fill the tank.

Then, the other tap will take (x + 3) hours to fill tank.

Total time taken to fill the tank by two taps = equals 3 1 over 13 space hours equals 40 over 13 space hours

Part filled by one tap in 1 hour equals 1 over x

Part filled by other tap in 1 hour equals fraction numerator 1 over denominator x plus 3 end fraction

Part filled by both the taps in 1 hour equals 13 over 40

Therefore, one tap takes 5 hours to fill the tank and other tap takes (5 + 3) 8 hours to fill the tank.

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 x 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 x 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 m

And length of the hall is (14 + 3) = 17 m

Solution 61

We know that,

Perimeter of a rectangle = 2(length + breadth)

⇒ 60 = 2(length + breadth)

⇒ length + breadth = 30 m

Let the length of a rectangular plot be x m.

Then, breadth of a rectangular plot = (30-x) m 

Now,

Area of a rectangular plot = 200 sq. m

Hence, the length and breadth of a rectangular plot are 20 m and 10 m respectively.

Solution 62

Let the width of the path be x m.

Length of the field including the path =   m

Breadth of the field including the path =   m

Area of the field including the path =

Area of the field excluding the path =   

Therefore, area of the path

Hence, the width of the path is 2 m.

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 24 m and 8 m 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.

Therefore, the dimensions of the triangle are 30 cm, 40 cm and 50 cm.

Solution 67

Let the altitude of triangle be x meter.

Hence, base = 3x meter

Hence, altitude of triangle is 8 m.

And base of triangle = 3x = (3 x 8) cm = 24 m

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 = 16 m

And the base of the triangle = (16 - 4) = 12 m

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

Solution 72

Let the speed of the fast train be x kmph.

Then, the speed of the slow train =   kmph 

Time taken by fast train to cover 200 km   hours

Time taken by slow train to cover 200 km   hours

Hence, the speeds of two trains are 50 kmph and 40 kmph respectively.

Solution 73

Let the speed of the stream be = x kmph

Speed of boat in still water = 18 kmph

Speed of boat upstream = (18-x) km/hr

Time taken by the boat to go 36 km upstream   hours

Speed of boat downstream = (18+x) km/hr

Time taken by boat to go 36 km downstream

Hence, the speed of the stream is 6 kmph.

Solution 74

Let the time taken by the second pipe to fill the tank = x hours

Then, the time taken by the first pipe to fill the tank = (x-10) hours

Total time taken by both pipes to fill the tank = 12 hours

Part filled by second pipe in 1 hour =

Part filled by first pipe in 1 hour =

Part filled by both pipes together in 1 hour =

Hence, the time taken by second pipe to fill the tank is 30 hours.

Solution 75

Let the time taken by smaller tap to fill the tank = x hours

Then, the time taken by larger tap to fill the tank = (x-2) hours

Total time taken by both pipes to fill the tank =

Part filled by smaller tap in 1 hour =

Part filled by larger tap in 1 hour =

Part filled by both taps together in 1 hour =

Hence, the smaller tap will take 3 hours and the larger tap will take 5 hours to fill the tank separately.

Solution 76

Let the original speed of the aircraft be x kmph.

Then, time taken to cover 600 km =   hours

Reduced speed = (x-200) kmph

Time taken to cover 600 km at this speed =   hours

Therefore, original speed of the aircraft = 600 kmph

And, original duration of the flight  

Solution 77

Let the time taken by larger pipe and the smaller pipe to fill the pool be   hours and   hours respectively.

Total time taken by both pipes to fill the tank = 12 hours

In 4 hours, part of pool filled by the larger pipe

In 9 hours, part of pool filled by the smaller pipe

Multiplying (i) by 4,

Subtracting (iii) from (ii),

Substituting y=30 in (i),

Hence, the larger pipe will take 20 hours and the smaller pipe will take 30 hours to fill the pool separately.

Quadratic Equations Exercise Test Yourself

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 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

Correct option: (b)

Given equation is  .

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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