Lesson: Quadratic Equations

EXERCISE 1

Question: 1

Check whether the following are quadratic

equations:

(i)

   

x 1 2 x 3  

(ii)

 

  

x 2x 2 3 x   

(iii)

     

x 2 x 1 x 1 x 3    

(iv)

    

x 3 2x 1 x x 5   

(v)

     

2x 1 x 3 x 5 x 1    

(vi)

 

x 3x 1 x 2   

(vii)

 

x 2 2x x 1  

(viii)

 

3 2

x 4x x 1 x 2    

Solution

(i) Given equation:

   

x 1 2 x 3  

Simplify the given equation.

x 1 2x 2x 6   

x 1 2x 2x 6 0    

x 1 6 0  

x 7 0 

x 0x. 7 0  

The resulting equation is of the form

ax bx c 0  

Hence, the given equation is a quadratic

equation.

(ii) Given equation:

 

  

x 2x 2 3 x   

Simplify the given equation.

x 2x 6 2x   

x 2x 2x 6 0   

x 4x 6 0  

The resulting equation is of the form

ax bx c 0  

Hence, the given equation is a quadratic

equation.

(iii) Given equation:

     

x 2 x 1 x 1 x 3    

Simplify the given equation.

       

x x 1 2 x 1 x x 3 1 x 3      

2 2

x x 2x 2 x 3x x 3      

x 2 2x 3 0    

3x 1 0  

3x 1 0 

The resulting equation is not of the form

ax bx c 0  

Hence, the given equation is not a quadratic

equation.

(iv) Given equation:

    

x 3 2x 1 x x 5   

Simplify the given equation.

     

x 2x 1 3 2x 1 x x 5    

2 2

2x x 6x 3 x 5x    

2 2

2x x x 6x 5x 3 0     

x 10x 3 0  

The resulting equation is of the form

ax bx c 0  

Hence, the given equation is a quadratic

equation.

(v) Given equation:

     

2x 1 x 3 x 5 x 1    

Simplify the given equation.

       

2x x 3 1 x 3 x x 1 5 x 1      

2 2

2x 6x x 3 x x 5x 5      

2 2

2x x 6x x x 5x 3 5 0       

x 11x 8 0  

The resulting equation is of the form

ax bx c 0  

Hence, the given equation is a quadratic

equation.

(vi) Given equation:

 

x 3x 1 x 2   

Simplify the given equation.

 

2 2

x 3x 1 x 4 2 x 2      

3x 1 4 4x 0   

7x 3 0 

The resulting equation is not of the form

ax bx c 0  

Hence, the given equation is not a quadratic

equation.

(vii) Given equation:

 

x 2 2x x 1  

Simplify the given equation.

 

3 3 3

x 2 3 x 2 x 2 2x 2x      

 

3 3

x 8 6x x 2 2x 2x 0     

3 2

x 6x 12x 2x 8 0     

3 2

x 6x 14x 8 0    

3 2

x 6x 14x 8 0   

The resulting equation is not of the form

ax bx c 0  

Hence, the given equation is not a quadratic

equation.

(viii) Given equation:

 

3 2

x 4x x 1 x 2    

Simplify the given equation.

 

3 2

x 4x x 1 x 2    

 

3 2 3 3 2

x 4x x 1 x 2 3 x 2 3 x 2          

2 2

4x x 1 8 6x 12x      

2 2

6x 4x 12x x 8 1 0     

The resulting equation is of the

2x 13x 9 0  

form .

ax bx c 0  

Hence, the given equation is a quadratic

equation.

Question: 2

Represent the following situations in the form of

quadratic equations:

(i) The area of a rectangular plot is .The

528m

length of the plot (in metres) is one more than

twice its breadth. We need to find the length

and breadth of the plot.

(ii) The product of two consecutive positive integers

is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The

product of their ages (in years) 3 years from

now will be 360. We would like to find Rohan’s

present age.

(iv) A train travels a distance of 480 km at a uniform

speed. If the speed had been 8 km/h less, then it

would have taken 3 hours more to cover the

same distance. We need to find the speed of the

train.

Solution

(i) Let the breadth of the plot metres

x

According to the question,

The length of the plot metres

2x 1 

Area of the rectangular plot

Length Breadth 

 

2x 1 x 528 

2x x 528 

2x x 528 0  

The resulting equation is of the form

ax bx c 0  

Hence, it is the desired representation of the

given situation in the form of quadratic

equation.

(ii) Let the first integer is and the second integer

is .

x 1

Product of the two integers

 

x x 1 

According to the question,

 

x x 1 306 

x x 306 

x x 306 0  

The resulting equation is of the form

ax bx c 0  

Hence, it is the desired representation of the

given situation in the form of quadratic

equation.

(iii) Let the age of Rohan is years.

According to the question,

The age of his mother years

x 26 

After 3 years,

The age of Rohan years

x 3 

The age of his mother

x 26 3  

x 29years 

Product of their age after 3 years

  

x 3 x 29  

According to the question,

  

x 3 x 29 360  

Simplify the equation.

   

x x 29 3 x 29 360   

x 29x 3x 87 360   

x 32x 360 87 0   

x 32x 273 0  

The resulting equation is of the form

ax bx c 0  

Hence, it is the desired representation of the

given situation in the form of quadratic

equation.

(iv) Let the speed of the train

xkm / h

From the formula of speed, distance and time,

Distance

Time

Speed



Time taken to cover 480 km hrs

480



Now, let the speed of the train km/h

 

x 8 

Time taken to cover 480 km hrs

480

x 8





According to the question, the train takes 3

hours more to cover the same distance at the

speed 8 km/h less than the uniform speed.

Thus,

480 480

x x 8

 



Simplify the above equation.

 

480

x 8 3 480

 

  

 

 

480 480

x 3 8 3 480

x x

   

   

   

   

3840

480 3x 24 480 0

    

3840

3x 24 0

  

3x 24x 3840 0  

The resulting equation is of the form

ax bx c 0  

Hence, it is the desired representation of the

given situation in the form of quadratic

equation.

EXERCISE 4.2 (6)

Question: 1

Find the roots of the following quadratic equations

by factorisation:

(i)

x 3x 10 0  

(ii)

2x x 6 0  

(iii)

2x 7x 5 2 0  

(iv)

2x x 0

  

(v)

100x 20x 1 0  

Solution

(i) Given equation:

x 3x 10 0  

Split the middle term to find out factors of the

equation.

x 3x 10 0  

x 5x 2x 10 0   

Take the common terms outside the

parenthesis.

   

x x 5 2 x 5 0   

  

x 5 x 2 0  

That is,

 

x 5 0 

 

x 2 0 

x 5

x 2 

Hence, and are the roots of the

x 5

x 2 

equation.

(ii) Given equation:

2x x 6 0  

Split the middle term to find out factors of the

equation.

2x x 6 0  

2x 4x 3x 6 0   

Take the common terms outside the

parenthesis.

   

2x x 2 3 x 2 0   

  

x 2 2x 3 0  

That is,

 

x 2 0 

 

2x 3 0 

x 2 



Hence, and are the roots of the

x 2 



equation.

(iii)

2x 7x 5 2 0  

Split the middle term to find out factors of the

equation.

2x 7x 5 2 0  

2x 5x 2x 5 2 0   

Take the common terms outside the

parenthesis.

   

x 2x 5 2 2x 5 0   

  

2x 5 x 2 0  

That is,

 

2x 5 0 

 

x 2 0 





x 2 

Hence, and are the roots of

 

x 2 

the equation.

(iv) Given equation:

2x x 0

  

We can rewrite as,

2x x 0

  

 

16x 8x 1 0

  

Split the middle term to find out factors of the

equation.

 

16x 8x 1 0

  

 

16x 4x 4x 1 0

   

Take the common terms outside the

parenthesis.

   

 

4x 4x 1 1 4x 1 0

   

  

4x 1 4x 1 0

  

That is,

 

4x 1 0 

 

4x 1 0 



Hence, and are the roots of the



equation.

(v)

100x 20x 1 0  

Split the middle term to find out factors of the

equation.

100x 20x 1 0  

100x 10x 10x 1 0   

Take the common terms outside the

parenthesis.

   

10x 10x 1 1 10x 1 0   

  

10x 1 10x 1 0  

That is,

 

10x 1 0 

 

10x 1 0 



Hence, and are the roots of the



equation.

Question: 2

Solve the problems given in Example 1.

Example statements:

(i) John and Jivanti together have 45 marbles. Both

of them lost 5 marbles each, and the product of

the number of marbles they now have is 124.

We would like to find out how many marbles

they had to start with.

(ii) A cottage industry produces a certain number of

toys in a day. The cost of production of each toy

(in rupees) was found to be 55 minus the

number of toys produced in a day. On a

particular day, the total cost of production was

Rs 750. We would like to find out the number of

toys produced on that day.

Solution

(i) Let John has marbles.

Number of marbles with Jivanti

45 x 

In case they both lose marbles,

Number of marbles with john

x 5 

Number of marbles with Jivanti

45 x 5  

40 x 

Product of marbles when they both lose 5

marbles

  

x 5 40 x  

According to the question,

  

x 5 40 x 124  

Simplify the equation,

   

x 40 x 5 40 x 124   

40x x 200 5x 124 0    

x 45x 200 124 0   

x 45x 324 0  

Split the middle term to find out factors of the

equation.

x 45x 324 0  

x 36x 9x 324 0   

Take the common terms outside the

parenthesis.

   

x x 36 9 x 36 0   

  

x 36 x 9 0  

That is,

 

x 36 0 

 

x 9 0 

x 36

x 9

Thus, John had or number of marbles.

If John had marbles, then

Jivanti had marbles.

45 36 9 

Now, if John had marbles, then

Jivanti had marbles.

45 9 36 

(ii) Let the industry produces number of toys in a

day.

According to the question,

The cost associated with the production of each

toy

55 x 

Total cost of production

 

x 55 x 

According to the question,

 

x 55 x 750 

Simplify the equation.

55x x 750 

x 55x 750 0  

Split the middle term to find out factors of the

equation.

x 55x 750 0  

x 25x 30x 750 0   

Take the common terms outside the

parenthesis.

   

x x 25 30 x 25 0   

  

x 25 x 30 0  

That is,

 

x 25 0 

 

x 30 0 

x 25

x 30

Hence, the number of produced toys are either

25 or 30.

Question: 3

Find two numbers whose sum is 27 and product is

182.

Solution

Let the first number be .

According to the question,

The second number be .

27 x

Product of the two numbers

 

x 27 x 

According to the question,

 

x 27 x 182 

Simplify the above equation.

27x x 182 

x 27x 182 0  

Split the middle term to find out factors of the

equation.

x 27x 182 0  

x 13x 14x 182 0   

Take the common terms outside the parenthesis.

   

x x 13 14 x 13 0   

  

x 13 x 14 0  

That is,

 

x 13 0 

 

x 14 0 

x 13

x 14

In case, the first number is 13.

The second number

27 13 

14

In case, the first number is 14.

The second number

27 14 

13

Hence, the desired numbers are and .

Question: 4

Find two consecutive positive integers, sum of whose

squares is 365.

Solution

Let the first positive integer is and the second

positive integer is .

x 1

Sum of the squares of the two numbers

 

x x 1  

According to the question,

 

x x 1 365  

Simplify the above equation.

2 2

x x 1 2x 365   

2x 2x 364 0  

x x 182 0  

Split the middle term to find out factors of the

equation.

x x 182 0  

x 14x 13x 182 0   

Take the common terms outside the parenthesis.

   

x x 14 13 x 14 0   

  

x 14 x 13 0  

That is,

 

x 14 0 

 

x 13 0 

x 14 

x 13

The positive integer is 13.

Thus, the first positive integer is 13.

The second positive integer

13 1 

14

Hence, the two consecutive positive integers are

and .

Question: 5

The altitude of a right triangle is 7 cm less than its

base. If the hypotenuse is 13 cm, find the other two

sides.

Solution

Let the base of right triangle’s be cm.

According to the question,

The right triangle’s altitude cm

 

x 7 

The right triangle’s hypotenuse cm

13

By the Pythagoras theorem,

     

2 2 2

Base Altitude Hypotenuse 

Put the values of base, altitude and hypotenuse.

   

2 2

x x 7 13  

Simplify the above equation.

2 2 2

x x 7 2 7 x 169     

2x 49 14x 169  

2x 14x 169 49 0   

2x 14x 120 0  

x 7x 60 0  

Split the middle term to find out factors of the

equation.

x 7x 60 0  

x 12x 5x 60 0   

Take the common terms outside the parenthesis.

   

x x 12 5 x 12 0   

  

x 12 x 5 0  

That is,

 

x 12 0 

 

x 5 0 

x 12

x 5 

The sides of a triangle cannot be negative.

Thus, .

x 12

Hence, the base of the right triangle is 12 cm.

The right triangle’s altitude .

12 7 5cm  

Question: 6

A cottage industry produces a certain number of

pottery articles in a day. It was observed on a

particular day that the cost of production of each

article (in rupees) was 3 more than twice the 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.

Solution

Let the industry produces number of articles in a

day.

According to the question,

The cost associated with the production of each toy

2x 3 

Total cost of production

 

x 2x 3 

According to the question,

 

x 2x 3 90 

Simplify the above equation.

2x 3x 90 0  

Split the middle term to find out factors of the

equation.

2x 3x 90 0  

2x 15x 12x 90 0   

Take the common terms outside the parenthesis.

   

x 2x 15 6 2x 15 0   

  

2x 15 x 6 0  

That is,

 

2x 15 0 

 

x 6 0 





x 6

Since the number of articles cannot be negative,

Therefore, .

x 6

Cost of each article

 

2 6 3  

12 3 

15

Hence, the industry produces 6 number of articles in

a day, and the cost of each article is Rs. 15.

EXERCISE 4.3 (11)

Question: 1

Find the roots of the following quadratic equations,

if they exist, by the method of completing the

square:

(i)

2x 7x 3 0  

(ii)

2x x 4 0  

(iii)

4x 4 3x 3 0  

(iv)

2x x 4 0  

Solution

(i) Given equation:

2x 7x 3 0  

Divide both sides by 2.

7 3

x x 0

2 2

  

7 3

x x

2 2

  

7 3

x 2 x

4 2

    

Add both sides.

 

 

 

2 2

7 7 7 3

x 2 x

4 4 4 2

   

     

   

   

7 49 3

4 16 2

 

  

 

 

7 49 24

4 16



 

 

 

 

7 25

4 16

 

 

 

 

7 5

4 4

 

  

 

 

7 5

4 4

 

That is,

7 5

4 4

 

7 5

4 4

 



x 3



Hence, and are the roots of the

x 3



equation.

(ii) Given equation:

2x x 4 0  

Divide both sides by 2.

x 2 0

  

x 2

 

x 2 x 2

   

Add both sides.

 

 

 

2 2

1 1 1

x 2 x 2

4 4 4

   

     

   

   

1 1

x 2

4 16

 

  

 

 

1 33

4 16

 

 

 

 

1 33

4 4

  

33 1

4 4

  

That is,

33 1





33 1

 



Hence, and are the roots of

33 1



33 1

 

the equation.

(iii) Given equation:

4x 4 3x 3 0  

Divide both sides by 4.

x 3x 0

  

3 3

x 2 x

2 4

    

Add both sides.

 

 

 

2 2

3 3 3 3

x 2 x

2 2 2 4

   

     

   

   

3 3 3

2 4 4

 

  

 

 

x 0

 

 

 

 

That is,

x 0

 

x 0

 

 

Hence, and are the roots of the



equation.

(iv) Given equation:

2x x 4 0  

Divide both sides by 2.

x 2 0

  

x 2 x 2

    

Add both sides.

 

 

 

2 2

1 1 1

x 2 x 2

4 4 4

   

     

   

   

1 1

x 2

4 16

 

  

 

 

1 1 32

4 16



 

 

 

 

1 31

4 16

 

  

 

 

The square of a number is always positive.

Hence, the given equation has no real root.

Question: 2

Find the roots of the quadratic equations given in

Q.1 above by applying the quadratic formula.

(i)

2x 7x 3 0  

(ii)

2x x 4 0  

(iii)

4x 4 3x 3 0  

(iv)

2x x 4 0  

Solution

(i) Given equation:

2x 7x 3 0  

Compare the given equation with

ax bx c 0  

To find the values of , and .

The values are .

a 2,

b 7, 

c 3

According to the quadratic formula,

b b 4ac

  



Put the values of , and .

   

7 7 4 2 3

2 2

      





7 49 24

 



7 25





7 5





For positive sign,

7 5





x 3

For negative sign,

7 5





Hence, and are the roots of the

x 3



equation.

(ii) Given equation:

2x x 4 0  

Compare the given equation with

ax bx c 0  

To find the values of .

The values are .

a 2,

b 1,

c 4 

According to the quadratic formula,

b b 4ac

  



Put the values of and .

   

1 1 4 2 4

2 2

     





1 1 32

  



1 33

 



For positive sign,

1 33

 



For negative sign,

1 33

 



Hence, and are the

1 33

 



1 33

 



roots of the equation.

(iii) Given equation:

4x 4 3x 3 0  

Compare the given equation with

ax bx c 0  

To find the values of , and .

The values are .

a 4,

b 4 3,

c 3

According to the quadratic formula,

b b 4ac

  



Put the values of and .

 

4 3 4 3 4 4 3

2 4

    





4 3 48 48

  



4 3 0

 



For positive sign,

4 3 0

 







For negative sign,

4 3 0

 







Hence, and are the roots of the









equation.

(iv) Given equation:

2x x 4 0  

Compare the given equation with

ax bx c 0  

To find the values of and .

The values are .

a 2,

b 1,

c 4

According to the quadratic formula,

b b 4ac

  



Put the values of and .

 

1 1 4 2 4

2 2

    





1 1 32

  



1 31

  



The square of a number is always positive.

Hence, the given equation has no real root.

Question: 3

Find the roots of the following equations:

(i)

x 3,x 0

  

(ii)

1 1 11

,x 4,7

x 4 x 7 30

   

 

Solution

(i) Given equation:

x 3

 

Simplify the given equation.

x 1 3x 

x 3x 1 0  

Compare the given equation with

ax bx c 0  

To find the values of , and .

The values are .

a 1,

b 3, 

c 1 

According to the quadratic formula,

b b 4ac

  



Put the values of , and .

     

3 3 4 1 1

2 1

       





3 9 4

 



3 13





For positive sign,

3 13





For negative sign,

3 13





Hence, and are the roots

3 13





3 13





of the equation.

(ii) Given equation:

1 1 11

x 4 x 7 30

 

 

Simplify the given equation.

   

  

x 7 x 4

x 4 x 7 30

  



 

  

x 7 x 4 11

x 4 x 7 30

  



 

  

11 11

x 4 x 7 30





 

  

x 4 x 7 30   

   

x x 7 4 x 7 30    

x 7x 4x 28 30    

x 3x 2 0  

Split the middle term to find out factors of the

equation.

x 3x 2 0  

x 2x x 2 0   

Take the common terms outside the

parenthesis.

   

x x 2 1 x 2 0   

  

x 2 x 1 0  

That is,

 

x 2 0 

 

x 1 0 

x 2

x 1

Hence, and are the roots of the

x 2

x 1

equation.

Question: 4

The sum of the reciprocals of Rehman’s ages, (in

years) 3 years ago and 5 years from now is . Find

his present age.

Solution

Let Rehman’s present age be years.

Rehman’s age three years ago

x 3 

Rehman’s age after five years

x 5 

Sum of the reciprocals of Rehman’s age three years

ago and after five years .

1 1

x 3 x 5

 

 

According to the question,

1 1 1

x 3 x 5 3

 

 

Simplify the above equation.

  

x 5 x 3 1

x 3 x 5 3

  



 

  

2x 2 1

x 3 x 5 3





 

    

3 2x 2 x 3 x 5   

6x 6 x 5x 3x 15    

x 2x 15 6x 6 0    

x 4x 21 0  

Split the middle term to find out factors of the

equation.

x 4x 21 0  

x 7x 3x 21 0   

Take the common terms outside the parenthesis.

   

x x 7 3 x 7 0   

  

x 7 x 3 0  

That is,

 

x 7 0 

 

x 3 0 

x 7

x 3 

Since age is always positive,

Thus, the present age of Rehman is years.

Question: 5

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 been 210. Find

her marks in the two subjects.

Solution

Let Shefali’s marks in Mathematics

x

Shefali’s marks in English

30 x 

By the condition, the product of Shefali’s marks in

Mathematics and English is .

210

  

x 2 30 x 3 210   

Simplify the above equation.

   

x 30 x 3 2 30 x 3 210     

30x x 3x 60 2x 6 210 0      

x 25x 156 0   

x 25x 156 0  

Split the middle term to find out factors of the

equation.

x 25x 156 0  

x 12x 13x 156 0   

Take the common terms outside the parenthesis.

   

x x 12 13 x 12 0   

  

x 12 x 13 0  

That is,

 

x 12 0 

 

x 13 0 

x 12

x 13

If Shefali’s marks in Mathematics are , then her

marks in English .

30 12 18  

If Shefali’s marks in Mathematics are ,

then her marks in English .

30 13 17  

Question: 6

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.

Solution

Let the length of the shorter side be metres.

The length of the larger side metres.

30 x 

Diagonal of a rectangle

   

2 2

length width 

Diagonal of the rectangle

   

2 2

x x 30  

According to the question,

   

2 2

x x 30 x 60   

Take square both sides.

     

2 2 2

x x 30 x 60   

       

2 2

2 2 2

x x 30 2x 30 x 60 2x 60     

2 2

2x x 60x 120x 900 3600 0     

x 60x 2700 0  

Split the middle term to find out factors of the

equation.

x 60x 2700 0  

x 90x 30x 2700 0   

Take the common terms outside the parenthesis.

   

x x 90 30 x 90 0   

  

x 90 x 30 0  

That is,

 

x 90 0 

 

x 30 0 

x 90

x 30 

Sides of a rectangle are always positive.

Thus,

The length of the shorter side is 90 metres and the

length of the larger side is metres.

30 90 120 

Question: 7

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.

Solution

Let the larger number be and the smaller number

be y.

The difference of squares of two numbers

2 2

x y 

According to the question,

)

2 2

x y 180 ......(1) 

Also,

y 8x ......(2)

Put in equation (1).

x 8x 180 

x 8x 180 0  

Split the middle term to find out factors of the

equation.

x 8x 180 0  

x 18x 10x 180 0   

Take the common terms outside the parenthesis.

   

x x 18 10 x 18 0   

  

x 18 x 10 0  

That is,

 

x 18 0 

 

x 10 0 

x 18

x 10 

Since, the square of the smaller number is 8 times

the larger number and the square cannot be

negative, thus the larger number cannot be negative.

Therefore, .

x 18

Put in equation (2),

y 8 18 

y 144

y 144 

y 12 

Hence the larger number is and the smaller

number is or .

12

12

Question: 8

A train travels 360 km at a uniform speed. If the

speed had been 5 km/h more, it would have taken 1

hour less for the same journey. Find the speed of the

train.

Solution

Let the train’s speed km/hr

x

From the formula of speed, distance and time,

Distance

Time

Speed



Time taken to travel 360 km hr

360



If the speed had been 5 km/h more, then the time

taken to travel 360 km

360

x 5





According to the question,

360 360

x 5 x

 

 

 



 

Simplify the above equation,

 

360

x 5 1 360

 

  

 

 

5 360

360 x 5 360 0



    

x 1800 5x 0   

x 5x 1800 0  

Split the middle term to find out factors of the

equation.

x 5x 1800 0  

x 45x 40x 1800 0   

Take the common terms outside the parenthesis.

   

x x 45 40 x 45 0   

  

x 45 x 40 0  

That is,

 

x 45 0 

 

x 40 0 

x 45 

x 40

Since, speed is always positive,

Thus,

The speed of the train is km/h.

Question: 9

Two water taps together can fill a tank in hours.

The tap of larger diameter takes 10 hours less than

the smaller one to fill the tank separately. Find the

time in which each tap can separately fill the tank.

Solution

Let the time taken by small tap be hr.

The time taken by larger tap hr

x 10 

Part of tank that small tap fills in one hour



Part of tank that larger tap fills in one hour

x 10





Time taken by both taps to fill the tank hr



According to the question,

1 1 8

x x 10 75

 



Simplify the above equation.

 

x 10 x 8

x x 10 75

 





   

75 2x 10 8x x 10  

150x 750 8x 80x  

8x 230x 750 0  

Split the middle term to find out factors of the

equation.

8x 230x 750 0  

8x 200x 30x 750 0   

Take the common terms outside the parenthesis.

   

8x x 25 30 x 25 0   

  

x 25 8x 30 0  

That is,

 

x 25 0 

 

8x 30 0 

x 25



If the time taken by small tap is , the time taken

by larger tap

30 30 80 50

8 8 8



    

Which is not possible because time is always

positive.

Thus,

The time taken by small tap is 25 hours and the time

taken by larger tap .

25 10 15hours  

Question: 10

An express train takes 1 hour less than a passenger

train to travel 132 km between Mysore and

Bangalore (without taking into consideration the

time they stop at intermediate stations). If the

average speed of the express train is 11km/h more

than that of the passenger train, find the average

speed of the two trains.

Solution

Let the passenger train’s average speed be km/h.

Express train’s average speed km/h

 

x 11 

Time taken by the passenger train to travel 132 km

132



Time taken by the express train to travel 132 km

132

x 11





According to the question,

132 132

x x 11

 



Simplify the above equation,

132 132

x x 11

 



 

x 11 x

132 1

x x 11

 

 



 



 

 

132 11

x x 11







 

132 11 x x 11  

x 11x 1452 

x 11x 1452 0  

Split the middle term to find out factors of the

equation.

x 11x 1452 0  

x 44x 33x 1452 0   

Take the common terms outside the parenthesis.

   

x x 44 33 x 44 0   

  

x 44 x 33 0  

That is,

 

x 44 0 

 

x 33 0 

x 44 

x 33

Since speed is always positive,

Thus,

The passenger train’s average speed is 33 km/h and

the express train’s average speed

33 11 44km / h  

Question: 11

Sum of the areas of two squares is . If the

468m

difference of their perimeters is , find the sides

24m

of the two squares.

Solution

Let the side of the first square be metres and the

side of second square be metres.

Perimeter of first square

4x

Perimeter of second square

4y

Difference of the perimeters of the two squares

4x 4y 

According to the question,

4x 4y 24 

x y 6 

x y 6 ......(1) 

Area of first square

x

Area of second square

y

Sum of the areas of the two squares

2 2

x y 

According to the question,

2 2

x y 468 

Put the value of from equation (1).

 

y 6 y 468  

Simplify the above equation.

 

2 2

y 6 2 y 6 y 468     

2y 12y 36 468 0   

2y 12y 432 0  

y 6y 216 0  

Split the middle term to find out factors of the

equation.

y 6y 216 0  

y 18y 12y 216 0   

Take the common terms outside the parenthesis.

   

y y 18 12 y 18 0   

  

y 18 y 12 0  

That is,

 

y 18 0 

 

y 12 0 

y 18 

y 12

Since the side of a square is always positive,

Thus,

y 12

From equation (1),

x 12 6 

x 18

Hence, the side of first square is metres and the

side of second square is metres.

EXERCISE 4.4 (5)

Question: 1

Find the nature of the roots of the following

quadratic equations. If the real roots exist, find them:

(i)

2x 3x 5 0  

(ii)

3x 4 3x 4 0  

(iii)

2x 6x 3 0  

Solution

(i) Given equation:

2x 3x 5 0  

Compare the given equation with the general

form of the quadratic equation

ax bx c 0  

to find the values of and .

The values are .

a 2,

b 3, 

c 5

Calculate the discriminant.

D b 4ac 

Put the values of and .

 

D 3 4 2 5    

D 9 40 

D 31 

Since ,

D 0

Thus,

The given equation has no real root.

(ii) Given equation:

3x 4 3x 4 0  

Compare the given equation with the general

form of the quadratic equation

ax bx c 0  

to find the values of and .

The values are .

a 3,

b 4 3, 

c 4

Calculate the discriminant.

D b 4ac 

Put the values of and .

 

D 4 3 4 3 4    

D 48 48 

D 0

Since ,

D 0

Thus, the given equation has real roots that are

equal.

To find the roots of the equation.

b D

 



Put the values of and .

 

4 3 0

2 3

  





2 0





Thus,

 

 

Hence, the roots of the equation are and .

(iii) Given equation:

2x 6x 3 0  

Compare the given equation with the general

form of the quadratic equation

ax bx c 0  

to find the values of and .

The values are .

a 2,b 6,c 3   

Calculate the discriminant.

D b 4ac 

Put the values of , and .

 

D 6 4 2 3    

D 36 24 

D 12

Since

D 0

Thus, the equation has distinct real roots.

Now, find the roots.

b D

 



Put the values of , and .

 

6 12

2 2

  





6 12





6 2 3





3 3





Thus,

3 3



 

3 3



 

Hence, the roots of the equation are and

3 3



3 3



Question: 2

Find the values of for each of the following

quadratic equations, so that they have two equal

roots.

(i)

2x kx 3 0  

(ii)

 

kx x 2 6 0  

Solution

(i) Given equations:

2x kx 3 0  

Compare the given equation with the general

form of the quadratic equation

ax bx c 0  

to find the values of and .

The values are .

a 2,

b k,

c 3

Calculate the discriminant.

D b 4ac 

Put the values of and .

D k 4 2 3   

… (1)

D k 24 

Since the equation has two equal roots,

D 0

From equation (1).

k 24 0 

k 24

k 24 

k 2 6 

Thus, .

k 2 6 

(ii) Given equations:

 

kx x 2 6 0  

Simplify the given equation.

kx 2kx 6 0  

Compare the given equation with the general

form of the quadratic equation

ax bx c 0  

to find the values of and .

The values are .

a k,

b 2k, 

c 6

Calculate the discriminant.

D b 4ac 

Put the values of and .

 

D 2k 4 k 6    

… (1)

D 4k 24k 

Since the equation has two equal roots,

D 0

From equation (1).

4k 24k 0 

 

4k k 6 0 

That is,

4k 0

k 6 0 

k 0

k 6

Since cannot be zero,

Thus, .

k 6

Question: 3

Is it possible to design a rectangular mango grove

whose length is twice its breadth, and the area is

? If so, find its length and breadth.

800m

Solution

Let mango grove’s breadth

l

Mango grove’s length

2l

Area of the mango grove

  

2l l

According to the question,

  

2l l 800

2l 800

800



l 400

l 400 0 ......(1) 

Compare the given equation with the general form of

the quadratic equation to find the

al bl c 0  

values of .

a,band c

The values are .

a 1,

b 0,

c 400 

Calculate the discriminant.

D b 4ac 

Put the values of and .

   

D 0 4 1 400    

D 1600

Since ,

D 0

The equation has real roots.

Thus, the rectangular mango grove can be designed.

From equation (1).

l 400 0 

l 400

l 400 

l 20 

Since the length is always positive,

Thus,

The mango grove’s breadth is metres and the

length metres.

2 20 40  

Question: 4

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.

Solution

Let the first friend’s age is years.

The second friend’s age years

 

20 x 

Four years ago:

The first friend’s age years.

 

x 4 

The second friend’s age

20 x 4  

16 x 

According to the question,

  

x 4 16 x 48  

Simplify the above equation.

  

x 4 16 x 48  

16x x 64 4x 48 0    

x 20x 112 0   

x 20x 112 0  

Compare the given equation with the general form of

the quadratic equation to find the

ax bx c 0  

values of and .

The values are .

a 1,

b 20, 

c 112

Calculate the discriminant.

D b 4ac 

Put the values of and .

 

D 20 4 1 112    

D 400 448 

D 48 

Since ,

D 0

The equation has no real root.

Thus, the given situation is not possible.

Question: 5

Is it possible to design a rectangular park of

perimeter 80 m and area ? If so, find its length

400m

and breadth.

Solution

Let the park’s length and breadth be and .

Perimeter of the park

 

2 l b 

According to the question,

 

2 l b 80 

l b

 

l b 40 

b 40 l ......(1) 

Area of the park

l b 

Put the value of from equation (1).

Area of the park

 

l 40 l  

According to the question,

 

l 40 l 400  

40l l 400 0  

l 40l 400 0  

Compare the given equation with the general form of

the quadratic equation to find the

al bl c 0  

values of and .

The values are .

a 1,

b 40, 

c 400

Calculate the discriminant.

D b 4ac 

Put the values of , and .

 

D 40 4 1 400    

D 1600 1600 

D 0

Since ,

D 0

The equation has equal and real roots.

Thus, the given situation is possible.

To find the roots of the equation.

b D

 



Put the values of and .

 

40 0

2 1

  





40 0





Thus,

 

 

20, 

20 

Thus, the roots of the equation are and .

Hence, the length of the park is metres and the

breadth of the park is metres.

40 20 20 