 EXERCISE 1
Question: 1
Check whether the following are quadratic
equations:
(i)
2
x 1 2 x 3
(ii)
2
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)
2
2
x 3x 1 x 2
(vii)
3
2
x 2 2x x 1
(viii)
3
3 2
x 4x x 1 x 2
Solution
(i) Given equation:
2
x 1 2 x 3
Simplify the given equation.
2
x 1 2x 2x 6 0
2
x 1 6 0
2
x 7 0
2
x 0x. 7 0
The resulting equation is of the form
.
2
ax bx c 0
Hence, the given equation is a quadratic
equation.
(ii) Given equation:
2
x 2x 2 3 x
Simplify the given equation.
2
x 2x 6 2x
2
x 2x 2x 6 0
2
x 4x 6 0
The resulting equation is of the form
.
2
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
.
2
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
2
x 10x 3 0
The resulting equation is of the form
.
2
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
2
x 11x 8 0
The resulting equation is of the form
.
2
ax bx c 0
Hence, the given equation is a quadratic
equation.
(vi) Given equation:
2
2
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
.
2
ax bx c 0
Hence, the given equation is not a quadratic
equation.
(vii) Given equation:
3
2
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
.
2
ax bx c 0
Hence, the given equation is not a quadratic
equation. (viii) Given equation:
3
3 2
x 4x x 1 x 2
Simplify the given equation.
3
3 2
x 4x x 1 x 2
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
2
2x 13x 9 0
form .
2
ax bx c 0
Hence, the given equation is a quadratic
equation.
Question: 2
Represent the following situations in the form of
(i) The area of a rectangular plot is .The
2
528m
length of the plot (in metres) is one more than
twice its breadth. We need to find the length
(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
2x 1 x 528
2
2x x 528
2
2x x 528 0
The resulting equation is of the form
.
2
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
x
is .
x 1
Product of the two integers
x x 1
According to the question,
x x 1 306
2
x x 306
2
x x 306 0
The resulting equation is of the form
.
2
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.
x
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
2
x 29x 3x 87 360
2
x 32x 360 87 0
2
x 32x 273 0
The resulting equation is of the form
.
2
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
x
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
3
x x 8
Simplify the above equation.
480
x 8 3 480
x
480 480
x 3 8 3 480
x x
3840
480 3x 24 480 0
x
3840
3x 24 0
x
2
3x 24x 3840 0
The resulting equation is of the form
.
2
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)
2
x 3x 10 0 (ii)
2
2x x 6 0
(iii)
2
2x 7x 5 2 0
(iv)
2
1
2x x 0
8
(v)
2
100x 20x 1 0
Solution
(i) Given equation:
2
x 3x 10 0
Split the middle term to find out factors of the
equation.
2
x 3x 10 0
2
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,
or
x 5 0
x 2 0
or
x 5
x 2
Hence, and are the roots of the
x 5
x 2
equation.
(ii) Given equation:
2
2x x 6 0
Split the middle term to find out factors of the
equation. 2
2x x 6 0
2
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,
or
x 2 0
2x 3 0
or
x 2
3
x
2
Hence, and are the roots of the
x 2
3
x
2
equation.
(iii)
2
2x 7x 5 2 0
Split the middle term to find out factors of the
equation.
2
2x 7x 5 2 0
2
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,
or
2x 5 0
x 2 0
or
5
x
2
x 2
Hence, and are the roots of
5
x
2
x 2
the equation.
(iv) Given equation:
2
1
2x x 0
8
We can rewrite as,
2
1
2x x 0
8
2
1
16x 8x 1 0
8
Split the middle term to find out factors of the
equation.
2
1
16x 8x 1 0
8
2
1
16x 4x 4x 1 0
8
Take the common terms outside the
parenthesis.
1
4x 4x 1 1 4x 1 0
8 1
4x 1 4x 1 0
8
That is,
or
4x 1 0
4x 1 0
or
1
x
4
1
x
4
Hence, and are the roots of the
1
x
4
1
x
4
equation.
(v)
2
100x 20x 1 0
Split the middle term to find out factors of the
equation.
2
100x 20x 1 0
2
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,
or
10x 1 0
10x 1 0
or
1
x
10
1
x
10 Hence, and are the roots of the
1
x
10
1
x
10
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
(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.
x
Number of marbles with Jivanti
45 x
In case they both lose marbles,
5
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
2
40x x 200 5x 124 0
2
x 45x 200 124 0
2
x 45x 324 0
Split the middle term to find out factors of the
equation.
2
x 45x 324 0
2
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,
or
x 36 0
x 9 0
or
x 36
x 9
Thus, John had or number of marbles.
36
9
36
45 36 9
Now, if John had marbles, then
9
45 9 36
(ii) Let the industry produces number of toys in a
x
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.
2
55x x 750
2
x 55x 750 0
Split the middle term to find out factors of the
equation.
2
x 55x 750 0
2
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,
or
x 25 0
x 30 0
or
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 .
x
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.
2
27x x 182
2
x 27x 182 0 Split the middle term to find out factors of the
equation.
2
x 27x 182 0
2
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,
or
x 13 0
x 14 0
or
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 .
13
14
Question: 4
Find two consecutive positive integers, sum of whose
squares is 365.
Solution
Let the first positive integer is and the second
x
positive integer is .
x 1
Sum of the squares of the two numbers
2
2
x x 1
According to the question,
2
2
x x 1 365
Simplify the above equation.
2 2
x x 1 2x 365
2
2x 2x 364 0
2
x x 182 0
Split the middle term to find out factors of the
equation.
2
x x 182 0
2
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,
or
x 14 0
x 13 0
or
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
13
and .
14
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.
x
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
2
x x 7 13
Simplify the above equation.
2 2 2
x x 7 2 7 x 169
2
2x 49 14x 169
2
2x 14x 169 49 0 2
x 7x 60 0
Split the middle term to find out factors of the
equation.
2
x 7x 60 0
2
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,
or
x 12 0
x 5 0
or
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
x
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.
2
2x 3x 90 0
Split the middle term to find out factors of the
equation.
2
2x 3x 90 0
2
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,
or
2x 15 0
x 6 0 or
15
x
2
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)
2
2x 7x 3 0
(ii)
2
2x x 4 0
(iii)
2
4x 4 3x 3 0
(iv)
2
2x x 4 0
Solution
(i) Given equation:
2
2x 7x 3 0
Divide both sides by 2.
2
7 3
x x 0
2 2
  2
7 3
x x
2 2
2
7
4
2 2
2
7 7 7 3
x 2 x
4 4 4 2
2
7 49 3
x
4 16 2
2
7 49 24
x
4 16
2
7 25
x
4 16
7 5
x
4 4
7 5
x
4 4
That is,
or
7 5
x
4 4
7 5
x
4 4
or
12
x
4
2
x
4
or
x 3
1
x
2 Hence, and are the roots of the
x 3
1
x
2
equation.
(ii) Given equation:
2
2x x 4 0
Divide both sides by 2.
2
x
x 2 0
2
2
x
x 2
2
2
1
x 2 x 2
4
 
2
1
4
2 2
2
1 1 1
x 2 x 2
4 4 4
2
1 1
x 2
4 16
2
1 33
x
4 16
1 33
x
4 4
33 1
x
4 4 That is,
or
33 1
x
4
33 1
x
4
Hence, and are the roots of
33 1
4
33 1
4
the equation.
(iii) Given equation:
2
4x 4 3x 3 0
Divide both sides by 4.
2
3
x 3x 0
4
2
3 3
x 2 x
2 4
2
3
2
2 2
2
3 3 3 3
x 2 x
2 2 2 4
2
3 3 3
x
2 4 4
2
3
x 0
2
That is, or
3
x 0
2
3
x 0
2
or
3
x
2
3
x
2
Hence, and are the roots of the
3
2
3
2
equation.
(iv) Given equation:
2
2x x 4 0
Divide both sides by 2.
2
x
x 2 0
2
2
1
x 2 x 2
4
 
2
1
4
2 2
2
1 1 1
x 2 x 2
4 4 4
2
1 1
x 2
4 16
2
1 1 32
x
4 16
2
1 31
x
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)
2
2x 7x 3 0
(ii)
2
2x x 4 0
(iii)
2
4x 4 3x 3 0
(iv)
2
2x x 4 0
Solution
(i) Given equation:
2
2x 7x 3 0
Compare the given equation with
.
2
ax bx c 0
To find the values of , and .
a
b
c
The values are .
a 2,
b 7,
c 3
2
b b 4ac
x
2a
Put the values of , and .
a
b
c
2
7 7 4 2 3
x
2 2
7 49 24
x
4 7 25
x
4
7 5
x
4
For positive sign,
7 5
x
4
12
x
4
x 3
For negative sign,
7 5
x
4
2
x
4
1
x
2
Hence, and are the roots of the
x 3
1
x
2
equation.
(ii) Given equation:
2
2x x 4 0
Compare the given equation with
.
2
ax bx c 0
To find the values of . The values are .
a 2,
b 1,
c 4
2
b b 4ac
x
2a
Put the values of and .
a,
b,
c
2
1 1 4 2 4
x
2 2
1 1 32
x
4
1 33
x
4
For positive sign,
1 33
x
4
For negative sign,
1 33
x
4
Hence, and are the
1 33
x
4
1 33
x
4
roots of the equation.
(iii) Given equation:
2
4x 4 3x 3 0 Compare the given equation with
.
2
ax bx c 0
To find the values of , and .
a
b
c
The values are .
a 4,
b 4 3,
c 3
2
b b 4ac
x
2a
Put the values of and .
a,
b,
c
2
4 3 4 3 4 4 3
x
2 4
4 3 48 48
x
8
4 3 0
x
8
For positive sign,
4 3 0
x
8
3
x
2
For negative sign,
4 3 0
x
8
3
x
2 Hence, and are the roots of the
3
x
2
3
x
2
equation.
(iv) Given equation:
2
2x x 4 0
Compare the given equation with
.
2
ax bx c 0
To find the values of and .
a,
b
c
The values are .
a 2,
b 1,
c 4
2
b b 4ac
x
2a
Put the values of and .
a,
b,
c
2
1 1 4 2 4
x
2 2
1 1 32
x
4
1 31
x
4
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)
1
x 3,x 0
x
 
(ii)
1 1 11
,x 4,7
x 4 x 7 30
Solution
(i) Given equation:
1
x 3
x
 
Simplify the given equation.
2
x 1 3x
2
x 3x 1 0
Compare the given equation with
.
2
ax bx c 0
To find the values of , and .
a
b
c
The values are .
a 1,
b 3,
c 1
2
b b 4ac
x
2a
Put the values of , and .
a
b
c
2
3 3 4 1 1
x
2 1
3 9 4
x
2 3 13
x
2
For positive sign,
3 13
x
2
For negative sign,
3 13
x
2
Hence, and are the roots
3 13
x
2
3 13
x
2
of the equation.
(ii) Given equation:
1 1 11
x 4 x 7 30
Simplify the given equation.
x 7 x 4
11
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
2
x 7x 4x 28 30 2
x 3x 2 0
Split the middle term to find out factors of the
equation.
2
x 3x 2 0
2
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,
or
x 2 0
x 1 0
or
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
1
3
his present age.
Solution
Let Rehman’s present age be years.
x
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
2
6x 6 x 5x 3x 15
2
x 2x 15 6x 6 0
2
x 4x 21 0
Split the middle term to find out factors of the
equation.
2
x 4x 21 0
2
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,
or
x 7 0
x 3 0
or
x 7
x 3
Since age is always positive,
Thus, the present age of Rehman is years.
7
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
2
30x x 3x 60 2x 6 210 0
2
x 25x 156 0
2
x 25x 156 0 Split the middle term to find out factors of the
equation.
2
x 25x 156 0
2
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,
or
x 12 0
x 13 0
or
x 12
x 13
If Shefali’s marks in Mathematics are , then her
12
marks in English .
30 12 18
If Shefali’s marks in Mathematics are ,
13
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.
x
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
2
x 60x 2700 0
Split the middle term to find out factors of the
equation.
2
x 60x 2700 0
2
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,
or
x 90 0
x 30 0
or
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
x
be y.
The difference of squares of two numbers
2 2
x y
According to the question,
)
2 2
x y 180 ......(1)
Also,
2
y 8x ......(2)
Put in equation (1).
2
x 8x 180
2
x 8x 180 0
Split the middle term to find out factors of the
equation.
2
x 8x 180 0
2
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,
or
x 18 0
x 10 0
or
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),
2
y 8 18
2
y 144
y 144
y 12
Hence the larger number is and the smaller
18
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
x
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
1
x 5 x
Simplify the above equation,
360
x 5 1 360
x
5 360
360 x 5 360 0
x
2
x 1800 5x 0
2
x 5x 1800 0
Split the middle term to find out factors of the
equation.
2
x 5x 1800 0
2
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,
or
x 45 0
x 40 0
or
x 45
x 40
Since, speed is always positive,
Thus,
The speed of the train is km/h.
40
Question: 9
Two water taps together can fill a tank in hours.
3
9
8
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.
x
The time taken by larger tap hr
x 10
Part of tank that small tap fills in one hour
1
x
Part of tank that larger tap fills in one hour
1
x 10 Time taken by both taps to fill the tank hr
75
8
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
2
150x 750 8x 80x
2
8x 230x 750 0
Split the middle term to find out factors of the
equation.
2
8x 230x 750 0
2
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,
or
x 25 0
8x 30 0
or
x 25
30
x
8 If the time taken by small tap is , the time taken
30
8
by larger tap
30 30 80 50
10
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.
x
Express train’s average speed km/h
x 11
Time taken by the passenger train to travel 132 km
hr
132
x Time taken by the express train to travel 132 km
hr
132
x 11
According to the question,
132 132
1
x x 11
Simplify the above equation,
132 132
1
x x 11
x 11 x
132 1
x x 11
132 11
1
x x 11
132 11 x x 11
2
x 11x 1452
2
x 11x 1452 0
Split the middle term to find out factors of the
equation.
2
x 11x 1452 0
2
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,
or
x 44 0
x 33 0
or
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
2
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
x
side of second square be metres.
y
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
2
x
Area of second square
2
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).
x
2
2
y 6 y 468
Simplify the above equation.
2
2 2
y 6 2 y 6 y 468
2
2y 12y 36 468 0
2
2y 12y 432 0
2
y 6y 216 0
Split the middle term to find out factors of the
equation.
2
y 6y 216 0
2
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, or
y 18 0
y 12 0
or
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
18
side of second square is metres.
12
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)
2
2x 3x 5 0
(ii)
2
3x 4 3x 4 0
(iii)
2
2x 6x 3 0
Solution
(i) Given equation:
2
2x 3x 5 0
Compare the given equation with the general
2
ax bx c 0
to find the values of and .
a,
b
c
The values are .
a 2,
b 3,
c 5 Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
D 3 4 2 5
D 9 40
D 31
Since ,
D 0
Thus,
The given equation has no real root.
(ii) Given equation:
2
3x 4 3x 4 0
Compare the given equation with the general
2
ax bx c 0
to find the values of and .
a,
b
c
The values are .
a 3,
b 4 3,
c 4
Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
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
x
2a
Put the values of and .
a,
b
D
4 3 0
x
2 3
2 0
x
3
Thus,
2
,
3
2
3
Hence, the roots of the equation are and .
2
3
2
3
(iii) Given equation:
2
2x 6x 3 0
Compare the given equation with the general
2
ax bx c 0
to find the values of and .
a,
b
c
The values are .
a 2,b 6,c 3
Calculate the discriminant.
2
D b 4ac
Put the values of , and .
a
b
c 2
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
x
2a
Put the values of , and .
a
b
c
6 12
x
2 2
6 12
x
4
6 2 3
x
4
3 3
x
2
Thus,
3 3
,
2
3 3
2
Hence, the roots of the equation are and
3 3
2
.
3 3
2 Question: 2
Find the values of for each of the following
k
quadratic equations, so that they have two equal
roots.
(i)
2
2x kx 3 0
(ii)
kx x 2 6 0
Solution
(i) Given equations:
2
2x kx 3 0
Compare the given equation with the general
2
ax bx c 0
to find the values of and .
a,
b
c
The values are .
a 2,
b k,
c 3
Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
D k 4 2 3
… (1)
2
D k 24
Since the equation has two equal roots,
D 0
From equation (1).
2
k 24 0
2
k 24
k 24 k 2 6
Thus, .
k 2 6
(ii) Given equations:
kx x 2 6 0
Simplify the given equation.
2
kx 2kx 6 0
Compare the given equation with the general
2
ax bx c 0
to find the values of and .
a,
b
c
The values are .
a k,
b 2k,
c 6
Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
D 2k 4 k 6
… (1)
2
D 4k 24k
Since the equation has two equal roots,
D 0
From equation (1).
2
4k 24k 0
4k k 6 0
That is,
or
4k 0
k 6 0 or
k 0
k 6
Since cannot be zero,
k
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.
2
800m
Solution
l
Mango grove’s length
2l
Area of the mango grove
2l l
According to the question,
2l l 800
2
2l 800
2
800
l
2
2
l 400
2
l 400 0 ......(1)
Compare the given equation with the general form of
the quadratic equation to find the
2
al bl c 0
values of .
a,band c
The values are .
a 1,
b 0,
c 400 Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
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).
2
l 400 0
2
l 400
l 400
l 20
Since the length is always positive,
Thus,
The mango grove’s breadth is metres and the
20
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.
x
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
2
16x x 64 4x 48 0
2
x 20x 112 0
2
x 20x 112 0
Compare the given equation with the general form of
the quadratic equation to find the
2
ax bx c 0
values of and .
a,
b
c
The values are .
a 1,
b 20,
c 112
Calculate the discriminant.
2
D b 4ac
Put the values of and .
a,
b
c
2
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
2
400m
Solution
Let the park’s length and breadth be and .
l
b
Perimeter of the park
2 l b
According to the question,
2 l b 80
80
l b
2
l b 40
b 40 l ......(1)
Area of the park
l b
Put the value of from equation (1).
b
Area of the park
l 40 l
According to the question, l 40 l 400
2
40l l 400 0
2
l 40l 400 0
Compare the given equation with the general form of
the quadratic equation to find the
2
al bl c 0
values of and .
a,
b
c
The values are .
a 1,
b 40,
c 400
Calculate the discriminant.
2
D b 4ac
Put the values of , and .
a
b
c
2
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
l
2a
Put the values of and .
a,
b,
D
40 0
l
2 1 40 0
l
2
Thus,
40
,
2
40
2
20,
20
Thus, the roots of the equation are and .
20
20
Hence, the length of the park is metres and the
20
breadth of the park is metres.
40 20 20