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Math on Rough Sheets

Mathinstructor.net

Class 10 CBSE Ncert Solutions Chapter 4

Quadratic Equations

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Table of Contents: Page No.

Exercise 4.1 3

Exercise 4.2 9

Exercise 4.3 15

Exercise 4.4 33


3

Exercise 4.1

( )( ) ( )

( ) ( )( )

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( )( ) ( )( )

( ) ( )

( ) ( )

( ) ( )

2

2

22

3 2

3 2

. Check whether the following are Quadratic Equations.

1 2 -3

- 2 -2 3-

- 2 1 -1 3

-3 2 1 5

2 -1 -3 5 -1

3 1 - 2

2 2 -1

- 4 -

(

1 - 2

)

i x x

ii x x x

iii x x x x

iv x x x x

v x x x x

vi x x x

vii x x x

viii x x x x

+ =

=

+ = +

+ = +

= +

+ + =

+ =

+ =

Question 1

3

( ) Solution i

( ) ( ) ( )2 2 2 2

2

2

We just check degree of equation. If, degree of equation is equal to 2 then only

it is a quadratic eq

{ }

uation.

x+1 = 2 x-3 a+b =a +2ab+b

x +1+2x= 2x-6

x +7 = 0

Degree of equation is 2.

⇒

⇒

Therefore, it is a Quadratic Equation.


4

( )

( )( )2

2

2

2

2 2 3

2 6 2

2 2 6 0

4 6 0

Degree of equation is 2. Therefore, it is a Quadratic Equation.

x x x

x x x

x x x

x x

− = − −⇒ − = − +⇒ − − + =⇒ − + =

Solution ii

( )

( )( ) ( ) ( )2 2

2 2

2 1 1 3

2 2 3 3 0

2 2 3 3 0

2 2 3 3 0

3 1 0

Degree of equation is 1. Therefore, it is not a Quadratic Equation.

x x x x

x x x x x x

x x x x x x

x x x x

x

− + = − +⇒ + − − = + − − =⇒ + − − − − + + =⇒ − − − + + =⇒ − + =

Solution iii

( )

( )( ) ( )2 2

2 2

2

3 2 1 5

2 6 3 5

2 6 3 5 0

10 3 0

Degree of equation is 2. Therefore, it is a quadratic equation.

x x x x

x x x x x

x x x x x

x x

− + = +⇒ + − − = +⇒ + − − − − =⇒ − − =

Solution iv


5

( )

( )( ) ( )( )2 2

2 2

2

2 1 3 5 1

2 6 3 5 5

2 7 3 5 5 0

11 8 0

Degree of Equation is 2. Therefore, it is a Quadratic Equation.

x x x x

x x x x x x

x x x x x

x x

− − = + −⇒ − − + = − + −⇒ − + − + − + =⇒ − + =

Solution v

( )

( ) ( )2 22 2 2

2 2

2 2

3 1 2 2

3 1 4 4

3 1 4 4 0

7 3 0

Degree of equation is 1. Therefore, it is not a Quadratic Equatio

{

n.

}x x x a b a ab b

x x x x

x x x x

x

+ + = − − = − +

⇒ + + = + −⇒ + + − + − =

⇒ − =

Solution vi

( )

( ) ( ) ( )( ) ( ) ( )( )

3 32 3 3

3 3 2

3 3

3 3 2

3 2

2 2 1 3

2 3 2 2 2 1

8 6 2 2 2

2 2 8 6 12 0

6 14 8 0

Degree of

( ) { }

Equation is 3.

( )

x x x a b a b ab a b

x x x x x

x x x x x

x x x x x

x x x

+ = − + = + + +⇒ + + + = −⇒ + + + = −⇒ − − − − − =⇒ − − − =

Solution vii

Therefore, it is not a quadratic Equation.


6

( )

( ) ( ) ( )( ) ( ) ( )

3 33 2 3 3

3 2 3 3

2 2

2

4 1 2 3

4 1 2 3 2 2

4 1 8 6 12

2 13 9 0

Degree of Equation is 2. Therefore, it is a Quadratic

{

Equat

}

i

x x x x a b a b ab a b

x x x x x x

x x x x

x x

− − + = − − = − − −⇒ − − + = − − −⇒ − − + = − − +⇒ − + =

Solution viii

on.

( ) ( )2

. Represent the following situations in the form of Quadratic Equations:

The area of rectangular plot is 528 m . The length of the plot in metres is one more than twice

it

s breadth.

Question 2

i

2

We need to find the length and breadth of the plot.

:

We are given that area of a rectangular plot is 528 .

Let breadth of rectangular plot be x metres

Length is one more than twice its bread

m

Solution

( )th. Therefore, length of rectangular plot is 2 1

metres

x +

( )2

2

Area of rectangle = length × breadth

528 2 1

528 2

2 528 0 which is a .

x x

x x

x x

⇒ = +⇒ = +⇒ + − = Quadratic Equation


7

( )

( )( )

2

2

The product of two consecutive numbers is 306. We need to find the integers.

:

Let two consecutive numbers be 1 ..It is given that 1 306

306

306 0 which is a

x and x

x x

x x

x x

++ =

⇒ + =⇒ + − =

ii

Solution

Quadrati .c Equation

( ) ( ) Rohan's mother is 26 years older than him. The product of their ages in years after

3 years will be 360. We would like to find Rohan's present age.

:

Let present age of Rohan = years

Let

x

iii

Solution

( )

( )( )2

2

present age of Rohan's mother = 26 years

Age of Rohan after 3 years = 3 years

Age of Rohan's mother after 3 years = 26 3 29 years

According to given condition :

3 29 360

29 3 87 360

x

x

x x

x x

x x x

x

++

+ + = +

+ + =⇒ + + + =⇒ + 32 273 0, which is a .x − = Quadratic Equation

( ) A train travels a distance of 480 km at uniform speed. If, the speed had been

8 km/hr less, then it would have taken 3 hours more to cover the same distance. We

need to find speed of the train.

iv

: Let speed of train be /

480Time taken by train to cover 480 hours

480If, speed had been 8km/hr less then time taken would be hours

8

x km h

kmx

x

=

−

Solution

8

According to given condition, if speed had been 8km/hr less then time taken would be

3 hours less.

480 4803

8x x⇒ = +

−

8

,

( )

8( )

( 8)

.

( )( )

x x

x

x x

x x

x x

x

x x

x x

⇒ − =−

⇒ =

⇒ × = −⇒ = −⇒ − − =

−

−

+−

− =

2

2

2

1 1480 3

480 3

480 8 3 8

3840 3 24

3 24 3840 0

Dividing equation by 3 we get

8 1280 0 which is a Quadratic Equation


9

Exercise 4.2Exercise 4.2Exercise 4.2Exercise 4.2

( )( )( )

( )

( )

2

2

2

2

2

. Find the roots of the following Quadratic Equations by factorization.

i 3 10 0

ii 2 6 0

iii 2 7 5 2 0

1iv 2 0

8

v 10

0 20 1

0

x x

x x

x x

x x

x x

− − =

+ − =

+ + =

− + =

− + =

Question 1

( )

( ) ( )( )( )

2

2

3 10 0

5 2 10 0

5 2 5 0

5 2 0

5, 2

x x

x x x

x x x

x x

x

− − =⇒ − + − =⇒ − + − =

⇒ − + =⇒ = −

Solution i

( )

( ) ( )( ) ( )

2

2

2 6 0

2 4 3 6 0

2 2 3 2 0

2 3 2 0

3, 2

2

x x

x x x

x x x

x x

x

+ − =⇒ + − − =⇒ + − + =

⇒ − + =

⇒ = −

Solution ii


10

( )

( ) ( )( ) ( )

2

2

2 7 5 2 0

2 5 2 5 2 0

2 5 2 2 5 0

2 5 2 0

5, 2

2

x x

x x x

x x x

x x

x

+ + =

⇒ + + + =

⇒ + + + =

⇒ + + =

−⇒ = −

Solution iii

( )

( ) ( )( )( )

( )

( ) ( )( )( )

2

2

2

2

2

2

12 0

8

16 8 10

816 8 1 0

16 4 4 1 0

4 4 1 1 4 1 0

4 1 4 1 0

1 1,

4 4

100 20 1 0

100 10 10 1 0

10 10 1 1 10 1 0

10 1 10 1 0

1 1,

10 10

x x

x x

x x

x x x

x x x

x x

x

x x

x x x

x x x

x x

x

− + =

− +⇒ =

⇒ − + =⇒ − − + =⇒ − − − =⇒ − − =

⇒ =

− + =⇒ − − + =⇒ − − − =⇒ − − =

⇒ =

Solution iv

Solution v


11

( )( )

( )

( ) ( )( )( )

( )

2

2

2

2

2

. Solve the problems given in Example 1.

The problems given in example 1 are

i 45 324 0

ii 55 750 0

45 324 0

36 9 324 0

36 9 36 0

36 9 0

9,36

55

x x

x x

x x

x x x

x x x

x x

x

x x

− + =− + =

− + =⇒ − − + =⇒ − − − =⇒ − − =⇒ =

− +

Question 2

Solution i

Solution ii

( ) ( )( ) ( )

2

750 0

25 30 750 0

25 30 25 0

25 30 0

30,25

x x x

x x x

x x

x

=⇒ − − + =⇒ − − − =⇒ − − =⇒ =

( )

. Find two numbers whose sum is 27 and product is 182.

:

Let first number be

Let second number be 27

According to given condition, the product of two numbers is 182

Therefore, we ca

x

x−

Question 3

Solution

( )2

2

n write 27 182

27 182

27 182 0

x x

x x

x x

− =

⇒ − =⇒ − + =


12

( ) ( )( )( )

2 14 13 182 0

14 13 14 0

14 13 0

14,13

x x x

x x x

x x

x

⇒ − − + =⇒ − − − =

⇒ − − =⇒ =

Therefore, the first number is equal to 14 13

And, second number is 27 27 14 13

or Second number 27 13 27 13 14

Therefore two numbers are 13 and 14.

or

x= − = − == − = − =

( )

( )22

. Find two consecutive positive integers, sum of whose squares is 365.

Let first number be

Let second number be 1

According to given condition, we have

1 365

x

x

x x

+

+ + =

Question 4

Solution

( )

( ) ( )( ) ( )

2 2 2

2 2

2

2

2

2

1 2 365

2 2 364 0

Dividing equation by 2, we get

182 0

14 13 182 0

14 13 14 0

14 13 0

13, 14

Therefore first number

{

13

}

a b a b ab

x x x

x x

x x

x x x

x x x

x x

x

+ = + +⇒ + + + =⇒ + − =

+ − =⇒ + − − =⇒ + − + =⇒ + − =⇒ = −

= We discard 14 because it is given that number is positive}.

Second number 1 13 1 14

Therefore two consecutive positive integers are 13 and 14 whose sum of squares is equal t .

{

o 365

x

−= + = + =


13

( )

. The altitude of right triangle is 7 cm less than its base. If, hypotenuse is 13 cm.

Find the other two sides.

:

Let base of triangle be

Let altitude of triangle be 7 cm

It is

x cm

x −

Question 5

Solution

( ) ( )2 22 2 2 2

2 2

2

given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem, we can say that

13 7 2

169 49 14

169 2 14 49

2

x x a b a b ab

x x x

x x

= + − + = + +⇒ = + + −⇒ = − +⇒

( ) ( )( )( )

2

2

2

14 120 0

Dividing equation by 2, we get

7 60 0

12 5 60 0

12 5 12 0

12 5

5,12

We discard x=-5 because length of side of triangle cannot be negative.

Therefore, base of triangle 12

x x

x x

x x x

x x x

x x

x

− − =

− − =⇒ − + − =⇒ − + − =⇒ − +⇒ = −

=( )Altitude of triangle 7 12 7 5

cm

x cm= − = − =


14

( ) . A cottage industry produces a certain number of pottery articles in a day.

It was observed on a particular day that cost of production of each article in rupees was

3 more than twice the

Question 6

number of articles produced on that day. If, the total cost of production

on that day was Rs. 90, find the number of articles produced and the cost of each article.

:

Let cost of production of e

Solution

ach article be

We are given total cost of production on that particular day 90

90Therefore, total number of articles produced that day

According to the given conditions, we have

902( 3)

Rs x

Rs

x

xx

x

=

=

= +

⇒180

3x

= +

( ) ( )( )( )

2

2

2

180 3

180 3

3 180 0

15 12 180 0

15 12 15 0

15 12 0

15, 12

Cost cannot be in negative. Therefore, we discard 12

Therefore, =Rs 15 which is the cost of production of eac

( )

h

xx

xx x

x x

x x x

x x x

x x

x

x

x

+⇒ =

⇒ = +⇒ − − =⇒ − + − =⇒ − + − =⇒ − + =⇒ = −

= −

article.

90 90Number of articles produced on that particular day 6

15x= = =


15

Exercise 4.3

( )( )( )( )

2

2

2

2

.

.

2 7 3 0

2 4 0

4 4 3 3 0

2 4 0

x x

x x

x x

x x

− + =+ − =

+ + =+ + =

1 Find the roots of the following quadratic equations if they exist by the

method of completing square

i

ii

iii

iv

( )2

2

2

2

2 7 3 0

First we divide equation by 2 to make coefficient of x equal to 1, we get

2 7 3 0

2 27 3

02 2

7 7We divide middle term of the equation by 2x, we get

2.2 4

We add and subtr

x x

x x

xx

x

x

− + =

− + =

⇒ − + =

=

Solution i

( )

2

2 22

2 222 2 2

7 7 3act square of from the equation 0, we get

4 2 2

7 3 7 70

2 2 4 4

7 7 3 70 2

2 4 2 4{ }

xx

xx

xx a b a b ab

− + =

− + + − =

⇒ − + + − = − = + −

27 3 490

4 2( )

16x⇒ − + − =


16

2

2

2

7 49 3

4 16 27 49 24

4 167 25

4 16

,

7 5

4 4

5 7 123

4 4 45 7 2 1

4 4

( )

( )

4 2

( )

x

x

x

x

x

x

⇒ − = −

−⇒ − =

⇒ − =

⇒ − = ±

⇒ = + = =

−= + = =

Taking Square root on both sides we get

And

13,

2x⇒ =

( ) 2

2

2

22

2 4 0

,

2 02

2 ,

1

2.2 4

1Now squaring it we get

4

,

1 12

2 4

x x

xx

Dividing middle term by x we get

x

x

xx

+ − =

+ − =

=

⇒ + + − −

Solution ii

Dividing equation by 2 we get

Following procedure of completing square we get

( )2

2 2 2

2

0 24

1 33

{ }

( ) 04 16

a b a b ab

x

= + = + +

⇒ + − =


17

,

1 33

4 4

33 1 33 1

4 4 4,

33 1 33 1

4 4 4

33 1 33 1, ,

4 4

x

x

x

x

+ = ±

−⇒ = − =

− − −= − =

− − −=

Taking square root on both sides we get

And

Therefore

( ) 2

2

2

2 2

2

2

4 4 3 3 0

Dividing equation by 4 to make coefficient of x equal to 1, we get

33 0

4

Following the procedure of completing square, we get

3 3 33 0

4 2 2

33

2

x x

x x

x x

x x

+ + =

+ + =

+ + + − =

⇒ + +

Solution iii

( )2

2 2 2

2

3 30 2

4 4

30

2 ,

30

2

3

2 . , .

3

{

,

}

(

)

a b a b ab

x

x

x

x

+ − = + = + +

⇒ + =

+ =

⇒ = −

= −

Taking square root on both sides we get

Quadratic equation has two roots Here both the roots have equal value

Therefore value of3

,2 2

−


18

( )2

2

2

2

2 22

22

2 4 0

.

4 0

2 2

2 02

,

1 12 0

2 4 4

12

2 4

x x

x

x x

xx

xx

xx

+ + =

+ + =

⇒ + + =

+ + + − =

⇒ + + + −

Solution iv

Dividing equation by 2 to make coefficient of equal to 1

Following the procedure of completing square we get

( )2

2 2 2

2

2

10 2

4

1 12 0

4 161 1 31

24 16 16

,

.

Ther

{ }

( )

( )

a b a b ab

x

x

= + = + +

⇒ + + − =

−⇒ + = − =

Taking square root on both sides right hand side does not exist because

square root of negative number does not exist

efore, this equation doesn not have any solution.

( )( )( )( )

2

2

2

2

. Find the roots of the following Quadratic Equations by applying

quadratic formula.

2 7 3 0

2 4 0

4 4 3 3 0

2 4 0

x x

x x

x x

x x

− + =+ − =

+ + =+ + =

Question 2

i

ii

iii

iv


19

( )

( ) ( )( )

2

2 2

2

2

2 7 3 0

2 7 3 0 0,

, .

4 : ,

2

7 7 4 2 3

4

7 49 24

4

x x

x x ax bx c

b b acx

a

x

x

− + =

− + = + + == = − =

− ± −=

± − −=

± −⇒ =

Solution i

Comparing quadratic equation with general form we get

a 2 b 7 and c 3

Putting these values in quadratic formula we get

7 25

4

7 25 7 25,

4 412 2

,4 4

13,

2

x

x

x

x

±⇒ =

+ −⇒ =

⇒ =

⇒ =

( )

( ) ( )( )

2

2 2

2

2

2 4 0

2 4 0 0,

2, 1 4.

4 ,

2

4

1 1 4 2 4

1 3

x x

x x ax bx c

a b c

b b acx

a

x

x

+ − =+ − = + + =

= = = −

− ± −=

− ± − −=

− ±⇒ =

Solution ii

Comparing quadratic equation with the general form we get

and


3

4

1 33 1 33,

4 4x

− + − −⇒ =


20

( )

( ) ( )( )

2

2 2

2

2

4 4 3 3 0

4 4 3 3 0 0,

4, 4 3 3.

4 ,

4

2

4 3 4 3 4 3

8

x x

x x ax bx c

a b c

b b acx

a

x

x

+ + =

+ + = + + =

= = =

− ± −=

− ± −=

⇒

Solution iii

Comparing quadratic equation with the general form we get

and


4 3 0

8

3

2

3 3The quadratic equation has two roots. Here, both the roots are equal. Therefore, ,

2 2

x

x

− ±=

⇒ = −

− −=

( )

( ) ( )( )

2

2 2

2

2

2 4 0

Comparing quadratic equation 2 4 0 with the general form 0, we get

2, 1 and 4.

4Putting these values in quadratic formula : , we get

2

1 1 4 2 4

4

1 31

4

x x

x x ax bx c

a b c

b b acx

a

x

x

+ + =+ + = + + =

= = =

− ± −=

− ± −=

− ± −⇒ =

Solution iv

2

But, square root of negative number is not defined.

Therefore, Quadratic Equation 2 4 0 has no solution.x x+ + =


21

( )

( )

. Find the roots of the following equations :

13, 0

1 1 11, 4,7

4 7 30

x xx

xx x

− = ≠

− = ≠ −+ −

Question 3

i

ii

( )

2

2

2

2 2

2

13 0

13

1 3

3 1 0

Comparing equation 3 1 0 with general form 0, we get

1, 3 and 1

4Using Quadratic formula, to solve equation, we get

2

x xx

x

xx x

x x

x x ax bx c

a b c

b b acx

a

− = ≠

−⇒ =

⇒ − =⇒ − − =

− − = + + == = − = −

− ± −=

Solution i

where

( ) ( ) ( )23 3 4 1 1

2

3 13

2

3 13 3 13,

2 2

x

x

x

± − − −=

±⇒ =

+ −⇒ =


22

( )

( ) ( )( ) ( )

( )( )2

2

2 2

1 1 11 4,7

4 7 307 4 11

4 7 30

11

4 7 30

30 7 4 28

3 2 0

Comparing equation 3 2 0 with general form 0, we get

1, 3 and 2.

Using quadratic for

1

m

1

xx x

x x

x x

x x

x x x

x x

x x ax bx c

a b c

− = ≠ −+ −

− − +⇒ =

+ −

⇒ =+ −

⇒ − = − + −⇒ − + =

− + = + + == −

−

= =

Solution ii

where

( ) ( ) ( )

2

2

4ula to solve equation, we get

2

3 3 4 1 2

2

3 1

23 1 3 1

, 2, 12 2

b b acx

a

x

x

x

− ± −=

± − −=

±⇒ =

+ −⇒ = =


23

( )

( )

. The sum of reciprocals of Rehman's ages in years 3 years ago and 5 years from

1now is . Find his present age.

3:

Let present age of Rehman= years

Age of Rehman 3 years ago 3 year

x

x= −

Question 4

Solution

( )

( ) ( )( ) ( )( ) ( )( ) 2

2

2

s

Age of Rehman after 5 years 5 years

According to the given condition :

1 1 1

3 5 3

5 3 1

3 5 3

3 2 2 (x 3 5 15)

6 6 2 15

4 21 0

x

x x

x x

x x

x x x

x x x

x x

= +

+ =− +

+ + −⇒ =

− +⇒ + = − + −⇒ + = + −⇒ − − =

( ) ( )( )

2 2

2

2

Comparing quadratic equation 4 21 0 with general form 0, we get

1, 4 and 21

4Using quadratic formula to solve Quadratic equation, , we get

2

4 4 4 1 21

2

4 16 84

24

x x ax bx c

a b c

b b acx

a

x

x

x

− − = + + == = − = −

− ± −=

± − − −=

± +⇒ =

±⇒ = 10

24 10 4 10

,2 2

7, 3

x

x

s

+ −⇒ =

⇒ = −

Age cannot be in negative. Therefore, we discard 3.

Therefore, present age of Rehman is 7 years.

x = −


24

. In a class test, the sum of Shefali's marks in Mathematics and English is 30.

Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their

marks would have be

Question 5

en 210. Find her marks in the two subjects.

:

Let Shefali's marks in Mathematics

Let Shefali's marks in English 30

If, she had got 2 marks more in Mathematics, her marks would be 2

If, sh

x

x

x

== −

= +

Solution

e had got 3 marks less in English, her marks in English would be 30 3 27x x= − − = −

( ) ( )2

2

2 2

According to given condition :

2 27 210

27 54 2 210

25 156 0

Comparing quadratic equation 25 156 0 with general form 0, we get

1, 25 and 156

Applying Quadratic Formula

x x

x x x

x x

x x ax bx c

a b c

b bx

+ − =⇒ − + − =⇒ − + =

− + = + + == = − =

− ±=2 4

, we get2

ac

a

−

( ) ( )( )225 25 4 1 156

25 625 624 25 12 2

25 1 25 1,

2 213, 12

Therefore, Shefali's marks in Mathematics = 13 or 12

Shefali's marks in English 30 30 13 17

Or

Shefali's marks in English 3

2

0 30 12 18

x

x

x

x

x

x

± − −=

± − ±⇒ = =

+ −⇒ =

⇒ =

= − = − =

= − = − =

( ) ( )Therefore her marks in Mathematics and English are 13, 17 or 12, 18 .


25

. The diagonal of a rectangular field is 60 metres more than the shorter side.

If, the longer side is 30 metres more than the shorter side, find the sides of the field.

:

Let shorter

Question 6

Solution

( ) ( )2 2 2

2 2 2

2

side of rectangle

Let diagonal of rectangle 60

Let longer side of rectangle 30

According to Pythagoras theorem, we can say that

60 30

3600 120 900 60

x metres

x metres

x metres

x x x

x x x x x

x

== +

= +

+ = + +⇒ + + = + + +⇒ −

2 2

60 2700 0

Comparing equation 60 2700 0 with standard form 0,

1, 60 2700

x

x x ax bx c we get

a b and c

− =

− − = + + == = − = −

( ) ( ) ( )

2

2

4Applying quadratic formula ,

2

60 60 4 1 2700

60 3600 10800 60 14400 60 120

2 2 260 120 60 120

,2 2

180 60, 90, 30

2 2Length cannot be in negative. Therefore, we ignore 30.

T f

2

here o

b b acx we get

a

x

x

x

x

− ± −=

± − − −=

± + ± ±⇒ = = =

+ −⇒ =

−⇒ = = −

−

re 90 which means length of shorter side 90 metres and

length of longer side 30 90 30 120 metres

Therefore, length of sides are and in .

x

x

= == + = + =

90 120 metres


26

The difference of squares of two numbers is 180. The square of the smaller number is 8

times the larger number. Find the two numbers.

:

Let smaller number x=

Question 7

Solution

( )

( )

2 2

2

Let larger number

We are given that 180 1

Also, we are given that square of smaller number is 8 times the larger number.

8 2

Puttin

y

y x

x y

=− =

⇒ =

( ) ( )2

2

2 2

g equation 2 1 , we get

8 180

8 180 0

Comparing equation 8 180 0 with general form 0,

1, 8 and 180.

in

y y

y y

y y ay by c we get

a b c

− =⇒ − − =

− − = + + == = − = −

( ) ( ) ( )

2

2

4Using quadratic formula, ,

2

8 8 4 1 180

218, 10

8 64 720 8 784 8 28

2 2 28 28 8 28

,2 2

18, 10

b b acy we get

a

y

yy

y

y

− ± −=

± − − −=

⇒ = −± + ± ±⇒ = = =

+ −⇒ =

⇒ = −

( )

{ }

( ) ( )

2

2

2

Using equation 2 to find smaller number :

8

8 8 18 144

12

,

8 8 10 80 No real solution for x

Therefore two numbers are 12,18 12,18

x y

x y

x

And

x y

or

=⇒ = = × =⇒ = ±

= = × − = −

−


27

. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr more,

it would have taken 1 hour less for the same journey. Find the speed of the train.

:

Let the speed o

Question 8

Solution

2

2

f the train /

If, speed had been 5km/hr more, train would have taken 1 hour less.

So, according to this condition, we have

360 3601

51 1

360 15

5360 1

( 5)

360

(

5

5

)

5

1

x km hr

x x

x x

x x

x x

x x

x x

=

= ++

⇒ − =+

+ −⇒ = +

⇒ × = +⇒ + − 800 0=

( )( )

2 2

2

2

Comparing equation 5 1800 0 general equation 0,

1, 5 1800

4Applying quadratic formula ,

2

5 5 4 1 1800

5 25 7200 5

2

7225

2 2

x x with ax bx c we get

a b and c

b b acx to solve equation we get

a

x

x

x

+ − = + + == = = −

− ± −=

− ± − −=

− ± + − ±⇒ = =

⇒5 85

25 85 5 85

,2 2

40, 45

peed of train cannot be in negative. Therefore, we discard 45

Therefore, speed of train 40 /

x

x

S x

km hr

− ±=

− + − −⇒ =

⇒ = −

= −

=


28

3 . Two water taps together can fill a tank in 9 hours. The tap of larger diameter takes

810 hours less than the smaller one to fill the tank separately. Find the time in which each

tap can

Question 9

separately fill the tank.

:

Let time taken by tap of smaller diameter to fill the tank

Let time taken by tap of larger diameter to fill the tank 10

It means that tap of smaller

x hours

x hours

== −

Solution

1diameter fills th part of tank in 1 hour. 1

1And, tap of larger diameter fills part of tank in 1 hour. 2

10

3 75When two taps are used together, they fill tank in 9 hours hours

( )

8

( )

8

x

thx −

= =

( ) ( )

( ) 2

2

2

2

2

8In 1 hour, they fill part of tank 3

75From , and , we can say that

1 1 8

10 7510 8

( 10) 75

75 2 10 8 10

150 750 8 80

8 80 150 750 0

8 230 7

( )

(

50 0

4 115 17

)

0

( )

0

th

x xx x

x x

x x x

x x x

x x x

x x

x x

+ =−

− +⇒ =

−⇒ − = −⇒ − = −⇒ − − + =

⇒ − + =⇒ − + =

1 2 3

2 2

2

Comparing equation 4 115 375 0 with general equation 0,

we get 4, 115 and 375.

4Applying quadratic formula , we get

2

x x ax bx c

a b c

b b acx

a

− + = + + == = − =

− ± −=


29

( ) ( ) ( )2115 115 4 4 375

115 13225 6000

8

115 7225

8115 85

8115 85 115 85 200 30

,

8

, 25, 3.758 8 8 8

x

x

x

x

x

± − −=

± −⇒ =

±⇒ =

±⇒ =

+ −⇒ = = =

Time taken by larger tap 10 3.75 10 6.25 hours

Time cannot be in negative. Therefore, we ignore this value.

Time taken by larger tap 10 25 10 15 hours

Therefore, time taken by larger tap is 15

x

x

= − = − = −

= − = − =

hours and time taken by smaller tap is 25 hours.

An express train takes 1 hour less than a passenger train to travel 132 km between

without taking into consideration the time they stop at intermediate Mysore and Bangalore

stations

Question 10

.

If, the average speed of the express train is 11 km/h more than that of the passenger train,

find the average speed of two trains.

Solution :

Let average speed of passenger train /

Let average speed of express train 11 /

132Time taken by passenger train to cover 132 km hours

132Time taken by express train to cover 132 km

11

x km hr

x km hr

x

x

== +

=

=+

hours


30

According to the given condition, we have

132 1321

11x x= +

+

( ) ( )2

2

2 2

1 1132 1

11

11132 1

( 11)

132 11 11

1452 11

11 1452 0

Comparing equation 11 1452 0 with general quadratic equation 0,

we get 1, 11 and 1452.

Applying Qu

x x

x x

x x

x x

x x

x x

x x ax bx c

a b c

⇒ − = +

+ −⇒ = +

⇒ = +⇒ = +⇒ + − =

+ − = + + == = = −

2 4adratic Formula ,

2

b b acx we get

a

− ± −=

( ) ( ) ( )211 11 4 1 1452

2

11 121 5808

2

11 5929

211 77

211 77 11 77

,2 2

33, 44

We discard 44 because speed cannot be in negative.

Therefore, speed of passenger train 33 km/hr

And, speed of expre

x

x

x

x

x

x

x

− ± − −=

− ± +⇒ =

− ±⇒ =

− ±⇒ =

− + − −⇒ =

⇒ = −= −

=ss train 11 33 11 44 km/hrx= + = + =


31

2 . Sum of areas of two squares is 468 m . If, the difference of their perimeters is 24

metres, find the sides of the two squares.

:

Let perimeter of first square

Let perimete

x metres=

Question 11

Solution

r of second square 24

Length of side of first square 4

24Length of side of second square

4

Area of first square =

{ }

x metres

xmetres

xmetres

side

= +

= =

+=

×

Perimeter of square 4 x length of side

22

22

4 4 16

24Area of second square

4

x x xside m

xm

= × =

+ =

22

2 2

2 2

2

2

2

2

2


24468

16 4

576 48468

16 16

576 48468

162 576 48 468 16

2 48 576 7488

2 48 6912 0

24 3456 0

Comparing equation 24 345

x x

x x x

x x x

x x

x x

x x

x x

x x

+ + =

+ +⇒ + =

+ + +⇒ =

⇒ + + = ×⇒ + + =⇒ + − =⇒ + − =

+ − 26 0 with standard form 0,

1, 24 and 3456.

ax bx c

we get a b c

= + + == = = −

2 4

Applying Quadratic Formula , 2

b b acx we get

a

− ± −=


32

( ) ( )( )224 24 4 1 3456

2

24 576 13824 24 14400 24 120

2 2 2

x

x

− ± − −=

− ± + − ± − ±⇒ = = =

24 120 24 120,

2 2x

− + − −⇒ =

48, 72x⇒ = −

Perimeter of square cannot be in negative. Therefore, we discard 72x = −

Therefore, perimeter of first square 48 metres=

And, Perimeter of second square 24 48 2472 metresx= + = + =

Its Perimeter 48Side of First square = = = 12 metres

4 4

Its Perimeter 72And, Side of second Square = = = 18 metres

4 4


33

Exercise 4.4

( )( )( )

( )

2

2

2

2

. Find the nature of the roots of the following quadratic equations. If the real roots exist,

find them.

2 3 5 0

3 4 3 4 0

2 6 3 0

2 3 5 0

Comparing this equation

i x x

ii x x

iii x x

x x

− + =

− + =− + =

− + =

Question 1

Solution i2with general equation 0, we get

2, 3 and 5.

ax bx c

a b c

+ + == = − =

( ) ( )( )

( )

22

2

Discriminant 4 3 4 2 5 9 40 31

Discriminant is less than 0 which means equation has no real roots.

3 4 3 4 0

b ac

x x

= − = − − = − = −

− + =Solution ii

2Comparing this equation with general equation 0,

we get 3, 4 3 4.

ax bx c

a b x and c

+ + =

= = − =

( ) ( )( )22

2

.

Discriminant = 4 4 3 4 3 4 48 48 0

Discriminant is equal to zero which means equations has equal real roots.

4Applying quadratic formula to find roots we get,

2

4 3 0 2 3 2

6 3 3

Because, equa

b ac

b b acx

a

x

− = − − = − =

− ± −=

±= = =

2 2tion has two equal roots, it means ,

3 3x =


34

( )

( ) ( )( )

2

2

22

2 6 3 0

Comparing equation with general equation 0, we get

2, 6, 3.

Discriminant 4 6 4 2 3 36 24 12

Value of discriminant is greater than zero. Therefore, equat

x x

ax bx c

a b and c

b ac

− + =+ + =

= = − =

= − = − − = − =

Solution iii

2

ion has distinct and real roots.

4Applying quadratic formula to find roots we get,

2

6 12 6 2 3 3 3

4 4 2

3 3 3 3,

2 2

b b acx

a

x

x

− ± −=

± ± ±= = =

+ −⇒ =

( )( ) ( )

2

. Find the value of k for each of the following quadratic equations, so that they have

two equal roots.

2 3 0

2 6 0

x kx

kx x

+ + =− + =

Question 2

i

ii

( ) 2

2 2

2 3 0

We know that quadratic equation has two equal roots only when the value of discriminant is equal

to zero.

Comparing equation 2 3 0 with general quadratic equation 0,

x kx

x kx ax bx c

+ + =

+ + = + + =

Solution i

( )( )2 2 2

2, 3.

Discriminant 4 4 2 3 24

Putting discriminant equal to zero, we get

we get

a b k and c

b ac k k

= = =

= − = − = −


35

2

2

24 0

24

24 2 6

k

k

k

− =⇒ =

⇒ = ± = ±

( ) ( )

( ) ( )( )

2

2 2

22 2

2 6 0

2 6 0

Comparing quadratic equation, 2 6 0 with general form 0,

we get , 2 6.

Discriminant 4 2 4 6 4 24

We know that two roots of quadrat

kx x

kx kx

kx kx ax bx c

a k b k and c

b ac k k k k

− + =⇒ − + =

− + = + + == = − =

= − = − − = −

Solution ii

( )2

ic equation are equal only if discriminant is equal to zero.

utting discriminant equal to zero, we get

4 24 0

4 6 0

0, 6

P

k k

k k

k

− =⇒ − =⇒ =

2

2

According to definition of quadratic equation , quadratic equation is the equation of the form

0, where 0.

Therefore, in equation 2 6 0, we cannot have 0. Therefore, we discard 0.

ax bx c a

kx kx k k

+ + = ≠

− + = = =So, =k 6

2

. Is it possible to design a rectangular mango grove whose length is twice its breadth,

and the area is 800 m . If so, find its length and breadth.

Question 3

:Solution

Let breadth of rectangular mango grove x metres=

Let length of rectangular mango grove 2 x metres=


36

2 2Area of rectangle = length × breadth 2 2 mx x x= × =

According to given condition we have 22 800x =

{ }

2

We can directly solve this equation, we get x=±20 .

Lets solve it using quadratic formula.

4Applying quadratic formula, to solve equation, we get

2

0 1600 4020

2 220, 20

We discard negati

b b acx

a

x

x

− ± −=

± ±= = = ±

⇒ = −

ve value of because breadth of rectangle cannot be in negative.

Therefore, = breadth of rectangle 20

Length of rectangle = 2 2 20 40

x

x metres

x metres

== × =

. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product

of their ages in years was 48.

Question 4

Solution

( )( )

:

Let age of first friend years

Let age of second friend 20 years

Four years ago, age of first friend 4 years

x

x

x

== −

= −

( ) ( )

( )( )2

2

2

Four years ago, age of second friend 20 4 16 years


4 16 48

16 64 4 48

20 112 0

20 112 0

x x

x x

x x x

x x

x x

= − − = −

− − =⇒ − − + =⇒ − − =⇒ − + =


37

( ) ( )( )

2 2

22

Comparing equation, 20 112 0 with general quadratic equation 0, we get

1, 20 112.

Discriminant 4 20 4 1 112 400 448 48 0

Discriminant is less than zero which means we have no

x x ax bx c

a b and c

b ac

− + = + + == = − =

= − = − − = − = − <

real roots for this equation.

Therefore, the give situation is not possible.

2 . Is it possible to design a rectangular park of perimeter 80 metres and area 400 m . If so,

find its length and breadth.

Let length of park metres

We are given area of rectangula

x=

Question 5

Solution

{ }

( )

r park

Therefore, breadth of park

Perimeter of rectangular park

We are given perimeter of rectangle

( )

x

x

x

=

= = ×

= + = +

=

2400 m

400metres Area of rectangle length breadth

4002 length breath 2 metres

80 metres

Therefore, we can write

Comparing equation, with general quadratic equation ,

we get

a=1, b=-40 and c=400

( )

.

x

x

x

x x

x x

x x x x

x

+ =

+⇒ =

⇒ + =⇒ − + =⇒ − + =

− + = + + =

2

2

2

2

2 2

4002 80

x

4002 80

2 800 80x

2 80 800 0

40 400 0

40 400 0 a b c 0


38

( ) ( ) ( )Discriminant =

Discriminant is equal to 0. Therefore, two roots of equation are real and equal which means

that it is possible to design a rectangular park of perimeter 80

− = − − = − =22b 4ac 40 4 1 400 1600 1600 0

2 metres and area 400 .

Using quadratic formula , 2

4 020

2Here, both the roots are equal to 20.

Therefore, length of rectangular park 20 metres

Breadth of rectangu

m

xa

x

− ± −=

±= =

=

2b b 4acto solve equation we get

0

400lar park 20 metres

x= =

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