X CH 4 Second Degree Equations

Chapter 4
Second Degree Equations

Prepared by
Cecilia Joseph
St. John De Britto’s, A.I.H.S,
Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 41 08 /10 /2020

Equations To view class Click

An equation is a mathematical statement consisting of

an equal symbol ‘ = ’ between two algebraic expressions or

an algebraic expression and a numeral.

Eg: 2x + 8 = x + 3

5y + 4 = 24

5z + 4 = 0

Above written equations are equations in one variable.

In the first equation variable is ‘ x ’

In the second equation variable is ‘ y ’

In the third equation variable is ‘ z ’

Square problems

Q) When each side of a square is increased by 1 metre, the

perimeter becomes 36 metres. What is the length of a side of

the original square? Perimeter Square
Ans) Side of the new square =
4 Perimeter = 4 x Side
= 36 Perimeter
4 =9m ∴ Side = 4

Side of the original square = 9 − 1

=8m

Q) When each side of a square is increased by 1 metre, the

area becomes 36 square metres. What is the length of a side

of the original square? Square
Ans) Side of the new square = √ Area Area = Side x side

= √36 = 6 m Side = √ Area

Side of the original square = 6 − 1

=5m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

Q3) A box is to be made by cutting off small squares from each

corner of a square of thick paper, and bending upwards.

The height of the box is to be 5 centimetres and volume is
1
2 litres.

What should be the length of a side of the square sheet ?

Ans)

Given volume of the box = 1 litres
2
1
= 2 x 1000 1 litre = 1000 cm3

= 500 cm3

Given height of the box = 5 cm

We know,

Volume of the box = Base area x Height
Volume
∴ Base area = Height

= 500
5

= 100cm2

So, side of the base = √100 =10 cm

∴ Length of a side of the 5 cm
square sheet = 10 + 5+ 5
= 20 cm 10 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

Solving above question using algebraic method

5 cm 5 cm 5 cm

(X- 10) cm

X cm

Let length of a side of the original square be ‘ x ’ cm.

Side of base of the box = (x − 10) cm

Base area = (x − 10)2 cm2

Volume of the box = Base area x Height

= (x − 10)2 x 5

Given volume of the box = 500 cm3

∴ (x − 10)2 x 5 = 500

(x − 10)2 = 500
5

(x − 10)2 = 100

x − 10 = √100

x − 10 = 10

x = 10 + 10

x = 20

∴ Length of side of the square sheet = 20 cm

T.B Page 81
Q2) A square ground has a 2 metre wide path all around it. The

total area of the ground and the path is 1225 square metres.
What is the area of the ground alone?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

Ans) 2m

Let the side of the square ground 2m

be ‘ x ’. Xm
2m
Then,
(X + 4) m
Side of large square =x+4 2m
including path 5 1225
5 245
Area of large square = (x + 4 )2 7 49

Given area of large square =1225 cm2 7
5x5x7x7
( x + 4 )2 = 1225 5 x 7 = 35

x + 4 = √1225
x + 4 = 35

x = 35 − 4
x = 31

So , side of the square ground = 31 m
∴ Area of the square ground = (31)2

= 961 m2

Assignment
T. B Page 81

Q1) When each side of a square was reduced by 2 metres, the
area became 49 square metres. What was the length of a side
of the original square?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 42 09/10 /2020

To view class Click

Answer of last class assignment

T.B Page 81

Q1) When each side of a square was reduced by 2 metres, the

area became 49 square metres. What was the length of a side

of the original square?

Ans) Let the side of the original square be ‘ x ’
Side of the reduced square = x − 2
Area of the square = (x − 2)2
Given , area of the square = 49 m2

∴ (x − 2)2 = 49

x − 2 = √49
x− 2 = 7

x=7+2
x=9
∴ Side of the original square = 9 m

Q) When each side of a square was increased by 8 metres each,
its area becomes 1225 square metres. Find the sides of the
first square ?

Ans) Let the side of the original square be ‘x’
Side of the increased square = x + 8
Area of the square = (x + 8)2
Given , area of the square = 1225 m2

∴ (x + 8)2 = 1225

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

x + 8 = √1225
x + 8 = 35

x = 35 − 8
x = 27
∴ Side of the original square = 27 m

T.B Page 81

Q3) The square of a term in the arithmetic sequence 2, 5, 8,..........

is 2500. What is its position ?

Ans) Given arithmetic sequence is 2, 5, 8, .........

(First term ) f = 2

(Common difference ) d = 5 − 2

=3

Let square of nth term be 2500

We know,
nth term of any arithmetic sequence is

X n = an + b
where a = d, b = f − d

So, (an + b)2 = 2500 a=d=3

an + b = √2500 = 50

3n + −1 = 50 b=f−d=2−3
3n − 1 = 50 = −1

3n = 50 + 1

3n = 51
51
n= 3

n = 17

∴ Square of the 17th term is 2500

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

Q4) 2000 rupees was deposited in a scheme in which interest is

compounded annually. After two years the amount in the

account was 2205 rupees.

What is the rate of interest?

Ans)

If an amount ‘P’ is deposited for a period of ‘n’ years at the

rate of interest ’r’ compounded annually, then

Amount after ‘n’ years = P (1 + r n
100
)

Given,

Principal amount , P = 2000 rupees

No: of years , n = 2

So , Amount after 2 years = 2000 (1 + r 2
100
)

Given , Amount after 2 years = 2205 rupees

∴ 2000 (1 + r 2 = 2205
100
)

(1 + r 2 = 2205 = 441
100 2000 400
)

√(1+ r ) = 441
100 400

(1+ r ) = 21
100 20
r 21
100 = 20 −1

r = 21 − 20
100 20 20
r 21 −20
100 = 20

r = 1
100 20
1
r = 20 x 100

r =5

∴ Rate of interest, r = 5 %

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

Q) A red square, two yellow rectangles of
the same height and a small blue
square are kept together. The width of
the yellow rectangles and the side of
the blue square are all 1 metre. The
total area of the entire figure is
100 square metres.
Find the length of a side of the red square .

Ans)
Let the side of the red coloured
square be ‘x’ .

Then,

Total area = x2 + x + x + 1

= x2 + 2 x + 1

= x2 + 2 x x x 1 + 12 ( x + y)2 =x2 + 2 x y + y2

= ( x + 1)2

Given total area = 100 Sq.m

∴ ( x + 1)2 = 100

x + 1 = √ 100 = 10
∴ x = 10 – 1 = 9
∴ Length of a side of the red square = 9 m

We can also see (x + 1) 2 = x2 + 2 x +1 by
rearranging the pieces of the figure as
given here.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 5

Assignment

Q) In the figure there is a red square and two yellow rectangles
attached to it. Side of the square and height of each
rectangles are ‘x’ meters. Width of each rectangle is ‘a’ meters.
What is the area of the small square added so that the
rearranged figure becomes a large square?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 43 12 /10 / 2020

To view class Click

Answer of last class assignment

Q) In the figure there is a red square and two
yellow rectangles attached to it. Side of the
square and height of each rectangles are ‘x’
meters. Width of each rectangle is ‘a’meters.

What is the area of the small square added
so that the rearranged figure becomes a
large square?
Ans)

Rearranging the pieces of the given figure as below we can see,

Side of the small square to be
added to get a large square is ‘a’
x + a and its area is ‘ a2 ’

x+a

Conclusion:
Area of the square in the question is x2 + ax + ax = x2 + 2ax
To complete the square, a small square of area ‘a2 ’ is added.
Area of large square = x2 + 2ax + a2
To convert this algebraic expression into a perfect square use
the identity or consider the area of large square with side ‘x + a’
That is x2 + 2ax + a2 = ( x + a)2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

‘ a2 ’ is the square of half the coefficient of ‘ x’ ( 1 x 2 2 = a2
2
a)

Completing the square method
To convert ‘ x2 + 2 a x ’ to the perfect square( x + a)2
add the square of half the coefficient of ‘x’. i.e ‘ a2 ’

x2 + 2ax + a2 = ( x + a)2

Q) One side of a rectangle is 2 metres longer than the other side

and its area is 224 square metres. What are the lengths of the

sides?

Ans)

Let the small side be ‘ x ’, then the large side is ‘ x+2 ’.
Area of the rectangle = l x b = x x (x+2)

Given, area of rectangle = 224 sq.m
∴ x (x+2) = 224

x2 + 2 x = 224 Coefficient of x = 2
Completing the square
Half of it = 2 =1
x2 + 2 x +12 = 224 + 12 2

3 225

( x + 1)2 = 224 +1 3 75

( x + 1)2 = 225 5 25

x + 1 = √225 5

x + 1 = 15 3x3 x5x5

x = 15 − 1 = 14 3 x 5 =15

So, Small side = x = 14m
Large side = x + 2 = 14 + 2 = 16m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

T.B Page 86

Q1) 1 added to the product of two consecutive even natural

numbers gives 289. What are the numbers?

Ans)

If numbers are x, x+2 , their product is x( x+2 ),

1 added to the product is x( x+2 ) + 1

Given , 1 added to the product = 289
∴ x( x+2 ) + 1 = 289

x2 + 2x + 1 = 289

( x+1)2 = 289

x + 1 = √289 17 x 17 = 289

x + 1 = 17 ∴ √289 = 17

x = 17 − 1 = 16 so, x + 2 = 16 + 2 =18
∴ The numbers are 16 and 18

Q) A rectangle is to be made with perimeter 100 metres and area

525 square metres. What should be the length of its sides?

Ans)
Previous Knowledge

For a rectangle, Perimeter, P = 2 ( l+b ) , Area = l x b

∴ 2(l+b)=P
P
l+b= 2

Length , l= P −b
2
P
Breadth , b= 2 −l

Given perimeter of rectangle, P = 100m
P
Let length be ‘x’ then, breadth = 2 −l

= 100 −x = 50 − x
2

Given area of rectangle = 525 Sq.m
x (50 − x) = 525

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

50x −x2 = 525

−x2 + 50x = 525

Multiplying the equation by −1 we have Coefficient of x = 50

x2 − 50x = −525

Completing the square Half of it = 50 = 25
2

x2 − 50x + 252 = −525 + 252
x2 − 50x + 252 = −525 + 252
x2 − 50x + 252 = −525 + 625

( x − 25)2 = 100 ( x − y)2 = x2 − 2 x y + y2

x − 25 = √100
x − 25 = 10

x = 10 + 25
x = 35

∴ Sides of the rectangle are, length = x = 35m
breadth = 50 − x = 50 −35 =15m

Assignment

Q) The difference of two positive numbers is 6. Their produt is 216.
Find the numbers.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 44 13 /10 / 2020

To view class Click

Answer of last class assignment

Q) The difference of two positive numbers is 6. Their product is

216. Find the numbers.
Ans) Given difference of two positive numbers is 6.

So if one number is ‘x’ , other number may be x+6 or x−6

Let one number be ‘ x ’ , and other number be ‘ x + 6 ’

Their product is x (x + 6 )

Given product of numbers = 216 Coefficient of x = 6
∴ x ( x + 6 ) = 216
x2 + 6x = 216 Half of it = 6 =3
2
Completing the square
Square of 3 = 32
( Adding 32 to both sides )

x2 + 6x + 32 = 216 + 32

( x + 3 )2 = 216 + 9

( x + 3 )2 = 225

x + 3 = √225

x + 3 = 15

x = 15 − 3 = 12

x = 12 , x + 6 = 12 + 6 = 18
∴ Two positive numbers are 12 and 18 .

T. B Page 86
Q 2) 9 added to the product of two consecutive multiples of 6

gives 729. What are the numbers?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

Ans)
Given two consecutive multiples of 6

So if one multiple of 6 is ‘x’ , other multiple may be x+6 or x−6

Let the two consecutive multiples of 6 be ‘ x ’ and ‘ x + 6 ’

Their product is ‘ x (x + 6 ) ’

9 added to their product is ‘ x(x + 6 ) + 9 ’

Given, 9 added to their product = 729
∴ x(x + 6 ) + 9 = 729

x2 + 6x + 9 = 729

x2 + 6x + 32 = 729

( x + 3 )2 = 729 3 729
3 243
x + 3 = √729 3 81
x + 3 = 27 3 27
39
x = 27 − 3
= 24 3

x = 24 , x + 6 = 30 3x3x3x3x3x3
∴ The numbers are 24 and 30 3 x 3 x 3 = 27

Q3) How many terms of the arithmetic sequence 5, 7, 9,..............,

must be added to get 140?

Ans) n(n+1) + bn ,Where a = d
Sum of n terms of an A.S = a 2 b = f −d

Given A. S is 5, 7, 9 , ...............

First term , f = 5

Common difference, d = 7 − 5 = 2

Let the sum of ‘n’ terms be 140

Sum = a n(n+1) + bn a=d=2
2 b= f − d = 5 − 2 = 3

=2 n(n+1) + 3n
2

= n(n+1) + 3n

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

= n2 + n + 3n Coefficient of n = 4
= n2 + 4n
So, Half of it = 4 =2
n2 + 4n = 140 2
Completing the square
n2 + 4n + 22 = 140 + 22 Square of 2 = 22
( n + 2 )2 = 144
n + 2 = √144

= 12
n = 12 − 2 = 10

∴ Sum of 10 terms is 140

Q5) An isosceles triangle has to be made like this.

HeightHeight

Base

The height should be 2 metres less than the base and the area

of the triangle should be 12 square metres.

What should be the length of its sides?

Ans)

Let the base is ‘ x ’, then height is ‘ x − 2 ’
1
Area of triangle = 2 bh

Area of triangle = 1 x(x −2)
2

Given, area of triangle = 12 Sq.m
1
∴ 2 x(x −2) = 12

x(x −2) = 24

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

x2 − 2x = 24 Coefficient of x = 2

Completing the square Half of it = 2 =1
x2 − 2x + 12 = 24 + 12 2
( x − 1 )2 = 25
x − 1 = √25 = 5 Square of 1 = 12
x =5+1=6

∴ Base = x = 6m , Height = x − 2 = 6 − 2 = 4m

We know “ In an isosceles triangle , perpendicular drawn from
the vertex joining the equal sides, bisects the opposite side.”

Since Δ ABC is isosceles, AD bisects BC A
BC 6 4m
BC = 6m, BD = 2 = 2 = 3, AD = 4m 6Dm 3m

By Pythagoras theorem

AB2 = 32 + 42

= 9 + 16 = 25

AB = √25 = 5 B 3m C

∴ Sides of the triangle are 6m, 5m, 5m.

Assignment

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 45 16 /10 / 2020
Answer of last class assignment
To view class Click

Ans)
Let AB = h meters
Given AC = 2.6m and BC = 1m

A By Pythagores theorem
h 2.6 m
h = √(2.6)2−12

= √(2.6+1) x( 2.6−1) a2 −b2 = (a+b)(a −b)

= √3.6+1.6

√= 36 x 16
10 10

= √36 x √16
√100
B 1m C 6x4
= 10

= 24
10

= 2.4 m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2
2.6 m
Let the foot of the rod be moved ‘x’ m from the wall.

i.e, Let CQ =x metre

Given , upper end slides the same length,
∴ AP = x metre

A

Consider ΔPBQ 2.4m x
PB = 2.4 − x
BQ = 1 + x P 2.6 m
PQ = 2.6
2.4 – x

By Pythagores theorem

PB2 + BQ2 = PQ2 B 1m C x Q
(2.4 − x)2 + (1 + x)2 = (2.6)2
1+x

( (2.4)2 − 2 x 2.4 x x + x2 ) + (12 + 2 x 1 x x + x2 ) = 6.76

(a + b)2 = a2 + 2ab +b2

(a −b)2 = a2 − 2ab +b2

( 5.76 − 4.8 x + x2 ) + (1 + 2 x + x2 ) = 6.76

2 x2 − 2.8x + 6.76 = 6.76

2 x2 − 2.8x + 6.76 − 6.76 = 0

2x2 – 2.8x = 0

Taking 2x outside, 2x(x – 1.4) = 0
∴ 2x = 0 or (x – 1.4) = 0

So, x = 0 or x = 1.4

x = 0 means the rod is not sliding and this is not possible.
∴ x = 1.4 m.

i.e, Foot of the rod further moved = 1.4m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

Two answers

Q) One side of a rectangle is 2 metres longer than the other

side and its area is 224 square metres. What are the

lengths of the sides?

Ans) Let the breadth be ‘ x ’, then the length is ‘ x+2 ’.

Area of the rectangle = l x b = x x (x+2)

Given, area of rectangle = 224 sq.m x
∴ x (x+2) = 224

x2 + 2 x = 224

Completing the square X+2

x2 + 2 x +12 = 224 + 12

( x + 1)2 = 224 +1

( x + 1)2 = 225

x + 1 = √225 We know,

15 x 15 = 225

∴ x + 1 = ±15 Also, −15 x−15 = 225

When x + 1 = 15 When x + 1 = −15
x = 15 −1 x = −15 − 1
x = 14 x =−16

Breadth is always positive
∴ Breadth = x = 14m

Length =x + 2 = 14 + 2 = 16m

Q) The product of a number and 4 more than that number gives
480. Find the numbers ?

Ans) Let the numbers be x, x + 4
Product of the numbers : x ( x + 4)
Given product of the numbers is 480

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

∴ x ( x + 4) = 480 We know,
x2 + 4 x = 480 22 x 22 = 484

Completing the square Also, −22 x−22 = 484
x2 + 4 x + 22 = 480 + 22
( x + 2)2 = 480 + 4
( x + 2)2 = 484
x + 2 = √484

∴ x + 2 = ±22

When x + 2 = 22 When x + 2 = −22
x = 22 −2 x = −22 − 2
x = 20 x = −24

When x = 20 , x + 4 = 20 + 4 When x = −24 , x + 4 = −24 + 4
= 24 = −20

∴ The numbers are 20 , 24 or −24 , −20

Assignment
T.B page 91
Q1) The product of a number and 2 more than that is 168, what

are the numbers?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 46 19 /10 / 2020

To view class Click

Answer of last class assignment

T.B page 91

Q1) The product of a number and 2 more than that is 168, what

are the numbers?

Ans) Let the numbers be x, x + 2

Product of the numbers : x ( x + 2)

Given product of the numbers is 168
∴ x ( x + 2) = 168
x2 + 2 x = 168

Completing the square
x2 + 2 x + 12 = 168 + 12

( x + 1)2 =168 + 1

( x + 1)2 = 169

x + 1 = √169 We know,
13 x 13 = 169

∴ x + 1 = ± 13 Also −13 x−13 = 169

When x + 1 = 13 or When x + 1 = −13

x = 13 −1 x = −13 − 1

x = 12 x = − 14

When x = 12, x + 2 = 12 + 2 When x = −14 , x + 2 = −14 + 2

= 14 = − 12

∴ The numbers are 12 , 14 or −14 , −12

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

Q2) Find two numbers with sum 4 and product 2.

Ans) 1st No: + 2nd No: = 4
Let first number be ‘ x ’ . 2nd No: = 4−x
Since sum = 4, second number is ‘ 4 − x ’

Given product is ‘ 2 ’ .
∴ x ( 4 − x) = 2

4 x − x2 = 2

Multiplying both sides with −1 we have

x2 − 4 x = −2

Completing the square

x2 − 4 x + 22 = −2 + 22

(x − 2)2 = −2 + 4 +
(x − 2)2 = 2 -

Taking square root on both sides

x − 2 = ±√2

When x − 2 = √2 When x − 2 = − √2

x = 2 + √2 x = 2 − √2

So,

First number = x = 2 + √2 or First number = x = 2 − √2

Second number = 4 − x Second number = 4 − x

= 4 − ( 2 + √2 ) = 4 − ( 2 − √2 )

= 4 − 2 − √2 = 4 − 2 + √2

= 2 − √2 = 2 + √2
∴ The numbers are 2 + √2 and 2 − √2

Checking Answer

Sum = 2 + √2 + 2 − √2 = 4

Produt = (2 + √2 )( 2 − √2 ) =22 −( √2 )2 = 4 − 2 = 2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

Q3) How many terms of the arithmetic sequence 99, 97, 95,.......

must be added to get 900 ?

Ans) Given A. S is 99, 97, 95, ..........

First term, f = 99
Common difference, d = 97 − 99 = −2

Given, Sum = 900

We Know, a = d = −2

Sum = a n(n+1) + bn b = f − d = 99 −(−2)
2 = 99 + 2
= 101
= −2 x n(n+1) + 101n
2 – n + 101n = 101n – n
= 100n
= − n(n+1) + 101n

= −n2 − n + 101n
= −n2 + 100n

So, −n2 + 100n = 900

Multiplying both sides with −1 we have

n2 − 100n = −900

Completing the square

n2 − 100n + 502 = −900 + 502
(n − 50)2 = −900 + 2500
(n − 50)2 = 1600

Taking square root on both sides
n − 50 = √1600
n − 50 = ± 40

When n − 50 = 40 or When n − 50 = −40

n = 40 + 50 = 90 n = −40 + 50 = 10

∴ Number of terms to be added is 90 or 10

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

Q4) A rod 28 centimetres long is to be bent to make a rectangle.

(i) Can a rectangle of diagonal 8 centimetres be made?

(ii) Can a rectangle of diagonal 10 centimetres be made?

(iii) How about a rectangle of diagonal 14 centimetres?

Calculate the lengths of the sides of the rectangles that

can be made.

Ans) Perimeter = 2( l + b )

Given perimeter = 28cm

∴ 2( l + b ) = 28
28
l+b = 2 = 14

Let the length be ‘ x ’ then breadth is 14 − x

(i) Let us assume diagonal is 8 cm

By Pythagoras theorem 8 14 − x

( 14 − x )2 + x2 = 82

142 − (2 x 14 x x ) + x2 + x2 = 82 x

196 − 28x + 2x2 = 64

2x2 − 28x + 196 = 64

2x2 − 28x + 196 − 64 = 0

2x2 − 28x + 132 = 0

Dividing each term by 2 we get,

x2 − 14x + 66 = 0

x2 − 14x = −66

Completing the square

x2 − 14x + 72 = −66 + 72

(x − 7)2 = −66 + 49

(x − 7)2 = −17

This is not possible because perfect square is always positive.
∴ We cannot construct a rectangle with diagonal 8 cm.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 5

ii) Let us assume diagonal is 10 cm 10 14 − x
x
By Pythagoras theorem

( 14 − x )2 + x2 = 102
142 − (2 x 14 x x ) + x2 + x2 = 102

196 − 28x + 2x2 = 100
2x2 − 28x + 196 = 100
2x2 − 28x + 196 − 100 = 0

2x2 − 28x + 96 = 0

Dividing each term by 2 we get,
x2 − 14x + 48 = 0
x2 − 14x = −48

Completing the square
x2 − 14x + 72 = −48+ 72
(x − 7)2 = −48 + 49
(x − 7)2 = 1

x − 7 = √1
x− 7 = ± 1

When x − 7 = 1 or When x − 7 = −1
x = −1 + 7 = 6
x = 1+ 7 = 8

When x = 8 , other side = 14 − 8 = 6
When x = 6 , other side = 14 − 6 = 8
∴ We can construct a rectangle with diagonal 10 cm.
Assignment
A rod 28 centimetres long is to be bent to make a rectangle.
(iii) How about a rectangle of diagonal 14 centimetres?

To know more about pythagores click
**************************

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 47 20 /10 / 2020

To view class Click

Answer of last class assignment

T.B page 91

Q) A rod 28 centimetres long is to be bent to make a rectangle.

(iii) How about a rectangle of diagonal 14 centimetres?

Ans)

Let us assume diagonal is 14 cm

By Pythagoras theorem 14 14 − x
x
( 14 − x )2 + x2 = 142
142 − (2 x 14 x x ) + x2 + x2 = 142

196 − 28x + 2x2 = 196
2x2 − 28x + 196 = 196
2x2 − 28x + 196 − 196 = 0

2x2 − 28x = 0

Dividing each term by 2 we get,
x2 − 14x = 0

x ( x − 14) = 0
∴

x =0 or x − 14 = 0

x = 14

Here ,length = x = 0 Here ,length = x = 14

This is not possible breadth = 14 − x = 14 − 14

=0

This is not possible

∴ We cannot construct a rectangle with diagonal 14cm.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Q) A man travelled 300 km in a car at a constant speed. If the
speed of the car increased by 10 km/hr , he would have
reached 1 hr earlier. Find the speed of the car.

Ans)

Distance Distance = Speed x Time

÷÷ Distance
Time =
xSpeed Time
Speed

Speed = Distance
Time

Let speed of the car be ‘ x ’ .

Given , distance travelled = 300km

Time = Distance
Speed

Time(t) = 300 300 = 30
x 10
300
30 = 10

Of fractions with same

When speed is increased by 10km/hr numerator, the one with

smaller denominator is

300 the larger number.
(x + 10)
Time(t) = 300/10 is larger than

300/30

Here , 300 is larger than 300
x x +10

Given , time difference is ‘ 1 hr ’
300 300
∴ x − x +10 =1

Cross multiplying we get,

300(x + 10) − 300x =1
x( x + 10 )

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

300x + 3000 − 300x = x2 + 10x

3000 = x2 + 10x

x2 + 10x = 3000

Completing the square

x2 + 10 x + 52 = 3000 + 52

( x + 5)2 = 3000 + 25 5 3025
( x + 5)2 = 3025 5 605
11 121
x + 5 = √3025

∴ x + 5 = ± 55 11
5x5 x 11x11

5 x 11 = 55

When x + 5 = 55 or When x + 5 = −55

x = 55 − 5 x = −55 − 5

x = 50 x = − 60

Since the speed is positive , x = 50
∴ The speed of the car = 50km/hr .

Equations and Polynomials

Examples of second degree polynomials:

p(x) = x 2 + 2x + 1 , p(x) = 2x 2 + 3x + 1

Consider polynomial p(x) = 2x 2 + 3x + 1 ,
For x = −1, p( −1 ) = 2x 2 + 3x + 1 ,
= 2(−1)2 + 3 x (−1) + 1

= 2 − 3 +1 = 3 − 3 = 0
2x 2 + 3x + 1 = 0 is a second degree equation

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Any second degree polynomial can be put in the form

p(x) = ax2 + bx + c

Finding the number x for which p (x) = 0,

p(x) = 0

So, ax2 + bx + c = 0

ax2 + bx = —c

Dividing the equation completely by ‘a’

aax2+ bx =—ca
a

x2 + bx = —c
a a

Completing the square Coefficient of x = b
a
22 b
bx b —c b Half of it = 2a
x2 + a + 2a = a + 2a
b
Adding Square of 2a

b 2 —c + b2 both sides
2a a 4a2
x+ =

=—c x 4a2 + b2 x a

a x 4a2

= a (—c x 4a + b2 )

a x 4a2

= (—c x 4a + b2 )
4a2

= b2 − 4ac
4a2

Taking square root on both sides

√x + b =± b2−4 ac
2a 4 a2

= ± √b2−4 ac

2a

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

x = —b ± √b2−4 ac
2a
2a

x = —b ± √b2−4 ac

2a

Standard form of second degree equation is
ax2 + bx + c = 0 where a ≠ 0

To get ax2 + bx + c = 0, we must take

x = —b ± √b2−4 ac

2a
a = Coefficient of x2
b = Coefficient of x
c = Constant

Q) Find the solutions of the second degree equation

2x 2 + 3x +1 =0

Ans) Given equation is 2x 2 + 3x +1 = 0

Here a = 2 , b = 3 , c = 1

b 2 – 4ac = 32 – 4 × 2 × 1 = 9 – 8 = 1

x= −b±√b2−4 ac

2a

= − 3±√ 1

2x2
− 3±1
= 4

x= −3+1 or x= − 3−1
4 4
−2 −4
= 4 = 4

= −1 = —1
2

**********************
Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 1

Chapter 4
Second Degree Equations

Online class – 48 22 /10 / 2020

To view class Click

T. B Page 97

Q(2)How many consecutive natural numbers starting from 1

should be added to get 300?

Ans)

Let number of consecutive natural numbers added be ‘ n ’.

So, 1 + 2 + 3 + ... + n = 300 (Given)

We know ,

Sum of first ‘n’ natural numbers = n(n+1)
2

∴ n(n+1) = 300
2

n(n+1) = 2 × 300

n2 + n = 600

n2 + n − 600 = 0

Comparing this with the standard form ax2 + bx + c = 0

we have, a = Coefficient of n2
a = 1, b = 1, c = −600 b = Coefficient of n
c = Constant

b 2 – 4ac = 12 – 4 × 1 ×(−600)

= 1+ 2400 = 2401

n= −b±√b2−4 ac 7 2401
7 343
2a 7 49

= −1 ±√ 2401 7

2x1 7 x 7x 7x 7
−1±49
= 2 7 x 7 =49

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 2

n= −1+ 49 or n= −1−49
2 2
48 −50
= 2 = 2

= 24 = —25

Here, since ‘n’ is the number of terms , it is always positive.
∴ Number of consecutive natural numbers added = 24

Q4) In writing the equation to construct a rectangle of specified

perimeter and area, the perimeter was wrongly written as

24 instead of 42 . The length of a side was then computed as

10 metres.

* What is the area in the problem?

* What are the lengths of the sides of the rectangle in the

correct problem?

Ans) Perimeter = 2( l + b )

Perimeter was wrongly written as 24 m

∴ 2( l + b ) = 24
24
l+b = 2 = 12

Given, length of a side was computed then as 10 m

So, breadth= 12 −10 = 2 m

Then , Area =l × b = 10 × 2 = 20 sq m

Given , correct perimeter = 42 m

∴ 2( l + b ) = 42
42
l+b = 2 = 21

Let the length be ‘ x ’ then breadth is 21 − x
We have , Area = 20 sq m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 3

∴ x (21 −x) = 20
21 x − x2 = 20

Multiplying both sides with −1 we have

x2 − 21 x = −20

x2 − 21 x + 20= 0

Comparing this with the standard form ax2 + bx + c = 0

we have, a = Coefficient of x2
a = 1, b =−21, c = 20 b = Coefficient of x
c = Constant

b 2 – 4ac = (−21)2 – 4 × 1 ×(20)

= 441 − 80 = 361

x= −b±√b2−4 ac

2a

= −(−21 )±√ 361

2x1 19x19 =361
21±19 ∴ √361=19
= 2

x= 21+19 or x= 21−19
2 2
40 2
= 2 = 2

= 20 =1

∴ Sides of the rectangle are
Length = 20m , Breadth = 1m

Q5) In copying a second degree equation to solve it, the term
without x was written as 24 instead of −24. The answers
found were 4 and 6. What are the answers of the correct
problem?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 4

Ans) Standard form of second degree equation is
ax2 + bx + c = 0 where a ≠ 0

Given, the term without x (Constant) was wrongly written

as 24 .
∴ ax2 + bx + 24 = 0

Given, the answers found were 4 and 6.

So,

Substituting the value x = 4 in the above equation we have,

a(4)2 + b(4) + 24 = 0

16a + 4b + 24 = 0

16a + 4b =− 24

Dividing by 4

4a + b =− 6 ................ (1)

Substituting the value x = 6 in the equation we have,

a(6)2 + b(6) + 24 = 0

36a + 6b + 24 = 0

36a + 6b =− 24

Dividing by 6

6a + b =− 4 ................ (2)

Subtracting equation (1) from (2) 6a + b =− 4 ..........(2)
4a + b =− 6 ..........(1)
(2) − (1) 2a = − 4 – (− 6)

2a = 2 = −4 + 6 = 2
2
a= 2 =1

Substituting value ‘ a=1 ’ in equation(1)

4 × 1 + b =− 6

4 + b =− 6

b =− 6 − 4

b =−10

Substituting the values a = 1, b =−10 and c= −24 ( correct value)

in the standard equation ax2 + bx + c = 0 we have,

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 5

1x2 + (−10)x + (−24) = 0
x2 −10x − 24 = 0
Here a = 1, b =−10, c = −24
b 2 – 4ac = (−10)2 – 4 × 1 ×(−24)
= 100 + 96 = 196

x= −b±√b2−4 ac

2a

= −(−10 )± √196

2x1
10±14
= 2

x= 10+ 14 or x= 10−14
2 2
24 −4
= 2 = 2

= 12 = −2

∴ The answers of the correct problem are 12 and −2

Assignment
Q) What number added to 6 gives its own square?

(Hint: If the number is taken as ‘x’ , then equation is

x+6 = x2)

Project Work
Write any 10 second degree equations, write the coefficients
and constant term. Find the solutions of these equations ,find
the sum and product of these solutions.
Can you find any relation between the coefficients, constant
term and the sum and product of the solutions ?

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Second Degree Equations 6

Hint :

Second Degree Coefficient Coefficient Constant Solutions Sum of Product of
Equation of x2 of x of solutions solutions

a b c equations

x2+5x+6=0 1 5 6 −2, −3 −2 +−3 =−5 −2 ×−3= 6

x2−2x−15=0 1 −2 −15 −3, 5 −3 + 5 = 2 −3× −5=15

2x2−14x+24=0 2 −14 24 4, 3 4 + 3 = 7 4 × 3 =12

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi