1
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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
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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.
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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
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( )
( )( ) ( )( )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.
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( )
( ) ( ) ( )( ) ( ) ( )
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
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( )
( )( )
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
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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
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( )
( ) ( )( ) ( )
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
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( )( )
( )
( ) ( )( )( )
( )
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
− =
⇒ − =⇒ − + =
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( ) ( )( )( )
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
−= + = + =
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( )
. 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= − = − =
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( ) . 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= = =
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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⇒ − + − =
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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
= + = + +
⇒ + − =
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,
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
−
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( )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
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( )
( ) ( )( )
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
Putting these values in quadratic formula we get
3
4
1 33 1 33,
4 4x
− + − −⇒ =
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( )
( ) ( )( )
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
Putting these values in quadratic formula we get
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+ + =
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( )
( )
. 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
± − − −=
±⇒ =
+ −⇒ =
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( )
( ) ( )( ) ( )
( )( )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
− ± −=
± − −=
±⇒ =
+ −⇒ = =
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( )
( )
. 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 = −
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. 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 .
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. 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
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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
=⇒ = = × =⇒ = ±
= = × − = −
−
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. 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
− ±=
− + − −⇒ =
⇒ = −
= −
=
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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
− + = + + == = − =
− ± −=
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( ) ( ) ( )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
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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= + = + =
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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
According to given condition, we have
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
− ± −=
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( ) ( )( )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
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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 =
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( )
( ) ( )( )
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
= = =
= − = − = −
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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=
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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
According to given condition, we have
4 16 48
16 64 4 48
20 112 0
20 112 0
x x
x x
x x x
x x
x x
= − − = −
− − =⇒ − − + =⇒ − − =⇒ − + =
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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
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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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