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Question-1

Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)
(ii) x2 – 2x = -2(3 – x)

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)

 

Solution:
(i) (x + 1)2 = 2(x – 3)

x2 + 2x + 1 = 2x – 6

x2 + 7 = 0

This is of the form ax2 + bx + c = 0

It is a quadratic equation.


(ii) x2 – 2x = -2(3 – x)

x2 – 2x = -6 + 2x

x2 – 2x -2x + 6 = 0

x2 – 4x + 6. This is of the form ax2 + bx + c = 0

It is a quadratic equation.


(iii) (x - 2)(x + 1) = (x - 1)(x + 3)

x2 – 2x + x - 2 = x2 + 3x – x - 3

– x – 2x +1 = 0

- 3x + 1 = 0

It is not of the form ax2 + bx + c

It is not a quadratic equation.

Question-2

 Check whether the following are quadratic equations:

(i) (x – 3) (2x + 1) = x (x + 5)

(ii) (2x – 1) (x – 3) = (x + 5) (x – 1) (iii) x2 + 3x + 1 = (x – 2)2
(iv) (x+2)3 = 2x(x2 – 1) (v) x3 – 4x2 – x + 1 = (x – 2)3

 

Solution:
(i)(x – 3) (2x + 1) = x (x + 5)

2x2 + x – 6x – 3 = x2 + 5x

x2 - 3 = 0

This is of the form ax2 + bx + c = 0

Hence it is a quadratic equation.


(ii)(2x – 1) (x – 3) = (x + 5) (x – 1)

2x2 – 6x – x + 3 = x2 – x + 5x – 5

2x2 – 7x + 3 - x2 –4x + 5 = 0

x2 – 11x + 8 = 0

This is of the form ax2 + bx + x = 0

Hence it is a quadratic equation.


(iii) x2 + 3x + 1 = (x – 2)2

x2 + 3x + 1 = x2 – 4x +4

7x – 3 = 0

This is not of the form ax2 + bx + c = 0

It is not a quadratic equation.


(iv) (x+2)3 = 2x(x2 – 1)

x2 + 23 + 3x2 × 2 + 3x × 4 = 2x3 – 2x

x3 + 8 + 6x2 + 12x = 2x3 – 2x

x3 – 6x2 – 14x – 8 = 0

Its not of the form ax2 + bx + c = 0

It is not a quadratic equation.

(v) x3 – 4x2 – x + 1 = (x – 2)3

x3 – 4x2 – x + 1 = x3 – 8 - 3x2 × 2 + 3x × 22

x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x

2x2 –13x + 1 = 0

This of the form ax2 + bx + c = 0

Hence it is a quadratic equation.

Question-3

Represent the following situations in the form of quadratic equations:
(i)The area of a rectangular plot is 528 m2. The 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.

Solution:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let l be length (in metres) and b be the breadth (in metres) of the rectangle.
Given
l = 1 + 2b …………………(1)
lb = 528 m2 ----- (2)
(Area of rectangle = length × breadth)

Substituting (1) in (2)
(1+2b)b = 528
b +2b2 = 528
2b2 + b – 528 = 0
This is the quadratic equation where the breadth is in metres.

(ii) The product of two consecutive positive integers is 306. We need to find the integers
Let the 2 consecutive positive integer be x, x + 1
x(x + 1) = 306
x2 + x - 306 = 0 is the quadratic equation where x is the smallest
positive integer

Question-4

Represent the following situations in the form of quadratic equations:
(i) 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.
(ii) 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) 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.
Let the Rohans present age be x
Rohans mothers present age = x + 26
After 3 years,Rohans age = x + 3
Rohans mothers age = x + 26 + 3 = x + 29
(x + 3)(x + 29) = 360 (given)
x2 + 3x + 29x + 87 = 360
x2 + 32x – 273 = 0 where x is Rohans present age (in years)
   
(ii) 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.
Let the speed of the train be x km /hr
Time taken =
If speed = (x – 8) km/h
Time taken for the same distance = hr
  - = 3 (Given)
480x – 480 (x – 8) = 3(x – 8)x
480x – 480x + 480 x 8 = 3x2 – 24x
3x2 – 24x – 3840 = 0
x2 – 8x - 1280 = 0 is the required quadratic equation where x is the speed of the train in km /hr.

Question-5

Find the roots of the following quadratic equations by factorization:
(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0
(iii) + 7x + 5 = 0 (iv) 2x2 – x + = 0

(v) 100x2 – 20x + 1 = 0  

Solution:
(i)x2 – 3x – 10 = 0
x2 – 5x + 2x - 10 = 0
x(x – 5) + 2(x – 5) = 0
(x – 5)(x + 2) = 0
x = -2, x = 5
Thus –2, 5 are the roots of the equation.

(ii) 2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(x + 2) (2x - 3) = 0
x = -2 (or) x = are the roots of the equation

(iii) + 7x + 5 = 0
 x2 + 5x + 2x + 5 = 0
x() + (x + 5) = 0
(x + )(x + 5) = 0
x = -, x = -
The roots are -, -.

(iv) 2x2 – x + = 0
16x2 – 8x + 1 = 0
16x2 – 4x – 4x + 1 = 0
4x(4x – 1) – (4x – 1) = 0
(4x – 1)2 = 0
x = ,
The root of the equation are , .

(v) 100x2 – 20x + 1 = 0
100x2 – 10x – 10x + 1 = 0
10x(10x – 1) –1 (10x – 1) = 0
(10x – 1)2 = 0
x = ,
The roots of the equation are, .

Question-6

 Solve the following situation mathematically:

(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 the number of marbles John had be x

Then the number of marbles Jivanti had = 45 – x.

The number of marbles left with John, when he lost 5 marbles = x – 5

The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5

                                                                                           = 40 – x

Therefore, their product = (x – 5)(40 – x)

                                 =40x – x2 – 200 + 5x

                                 = -x2 + 45x – 200

So, -x2 + 45x – 200 = 124

i.e., -x2 + 45x – 324 = 0

i.e., x2 - 45x + 324 = 0

Therefore, the number of marbles John had, satisfies the quadratic equation

x2 – 45x + 324 = 0

Which is the required representation of the problem mathematically.

Now by solving we get, x2 – 9x – 36 x + 324 = 0

x(x – 9) - 36(x – 9) = 0

(x - 9)(x - 36) = 0

x = 9 (or) 36.

The number of marbles John had is 9 (or) 36.


(ii) Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy that day = 55 – x

So, the total cost of production (in rupees) that day = x(55 – x)

Therefore, x(55 – x) = 750

i.e., 55x – x2 = 750

i.e., -x2 + 55x – 750 = 0

i.e., x2 - 55x + 750 = 0

Therefore, the number of toys produced that day satisfies the quadratic equation

x2 – 55x + 750 = 0

which is the required representation of the problem mathematically.

Now by solving we get, x2 – 25x – 30x + 750 = 0

x(x – 25)- 30(x – 25) = 0

(x - 25)(x - 30) = 0

x = 25, x = 30.

The number of toys produced in a day = 25 (or) 30.

Question-7

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

Solution:
Let the two numbers be x and y
x + y = 27 …………………….. (1)
xy = 182 …………………….. (2)
from (1) x = 27 – y
Substituting in (2) we get,
(27 – y)y = 182
27y – y2 = 182
y2 – 27y + 182 = 0
y2 – 13y - 14y + 182 = 0
y(y – 13) – 14(y – 13) = 0
(y – 14)(y - 13) = 0
y = 14 (or) y = 13
If y = 14, x = 27 – 14 = 13
If y = 13, x = 27 – 13 = 14
Hence the two numbers are 13 and 14.

Question-8

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

Solution:
Let the 2 consecutive positive nos be x, x+1

x2 + (x + 1)2 = 365

x2 + x2 + 2x + 1 = 365

2x2 + 2x – 364 = 0

x 2 + x – 182 = 0 (Dividing by 2)

x2 + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x – 13)(x + 14) = 0

x = 13 (or) x = -14.

Since the positive integers cannot be -14

Hence the positive integers are 13 and 14.


Question-9

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:
In the right angled triangle
Let (BC) the base be x cm and altitude (AB) be (x – 7) cm
In the right angled triangle,
(AB)2 + (BC)2 = (AC)2 (Pythagoras theorem)
(x - 7)2 + x2 = 132
x2 – 14x + 49 + x2 = 169
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0 (Dividing by 2 )
x2 – 12x +5x - 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0
=> x = -5 (or) x = 12
Since the side (base) cannot be negative, x -5
We assume x = 12
=> Base (BC) = 12 cm
Altitude = x – 7 = 5 cm
Thus the two sides of the triangle are 5 cm and 12 cm.

Question-10

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 `90, find the number of articles produced and the cost of each article.    

Solution:
Let the number of articles be x and the cost of each article be y
Given:
y = 2x + 3 ………………..(1)
Total cost of production be xy = 90 …………….. (2)
Substuting y = 2x + 3 in (2), we get
x(2x + 3) = 90
2x2 + 3x – 90 = 0
2x2 + 15x – 12x – 90 = 0
2x2 + 15x – 12x – 90 = 0
x(2x + 15) - 6(2x + 15) = 0
(2x + 15) (x – 6) = 0
2x + 15 = 0 ; x – 6 = 0
=>x = ; 6
Since the number of articles cannot be negative as well as a fraction x
Hence the number of articles is 6 (i.e.,) x = 6
y = 2x + 3 from (1)
= 2(6) + 3 = 15
(i.e.,) Number of articles = 6
Cost of each article = `15

Question-11

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x + 3 = 0 (ii) 2x2 + x – 4 = 0
(iii) 4x2 +4x + 3 = 0 (iv) 2x2 + x + 4 = 0  

Solution:
(i)2x2 – 7x + 3 = 0
Dividing by the coefficient of x2, we get
x2 -
Comparing the quadratic equation ax2 + bx + c = 0
Here, a = 1, b = 7/2 and c = 3/2
Adding and subtracting the square of (b/2) =  we get,

x2 – 2 ´ × x + = 0
= 0
= 0
= 0

Taking the square root we get,

When
x = =
When
(or) x =

(ii) 2x2 + x – 4 = 0
Dividing by 2 (coefficient of x2), we get
x2 + = 0
Adding and subtracting the square of(b/2) =  we get,
x2 + 2(x) × +


= 0

Taking the square root we get,

x =
x = ;

(iii) 4x2 +4x + 3 = 0
(2x)2 + 2(2x) × + ()2 = 0
(2x +)2 = 0
2x = -
x = -
Roots of equation are x = -

(iv) 2x2 + x + 4 = 0
x2 + x/2 + 2 = 0
Adding and subtracting the square of
(b/2) = we get,
 
x2 + 2(x) ×

= 0

The square of a number cannot be negative. Hence roots do not exist.

Question-12

Find the roots of the following quadratic equations by applying the quadratic formula.
(i) 2x2 – 7x + 3 = 0 (ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4 = 0 (iv) 2x2 + x + 4 = 0

Solution:
(i) 2x2 – 7x + 3 = 0
According to the quadratic formula: ax2 + bx + c = 0
a = 2
b = -7
c = 3 The roots are =
                      =
                      = =
                      =
= = 3 ;
∴ The roots are 3, .

(ii) 2x2 + x – 4 = 0
According to the quadratic formula: ax2 + bx + c = 0
a = 2, b = 1, c = -4
The roots are
                       =
                  =
The roots are ,
 
(iii) 4x2 + 4 = 0
According to the quadratic formula: ax2 + bx + c = 0
a = 4, b = 4, c = 3
The roots are =
                      =
                      =
                      =
                      = ,
The roots are (,=  

(iv)2x2 + x + 4 = 0
According to the quadratic formula: ax2 + bx + c = 0
a = 2, b = 1, c = 4
The roots are =
                      =
                      =
                      =
This is not possible, hence the roots do not exists.

Question-13

Find the roots of the following equations:
(i) , x 0
(ii) , x -4, 7  

Solution:
(i) , x 0
x2 – 1 = 3x
x2 – 3x – 1 = 0
a = 1, b = -3, c = -1
The roots are x =
x =
   =
(i.e.,) x = and
 
(ii)



(x+4)(x – 7) = -30
x2 – 3x – 28 = -30
x2 – 2x - x + 2 = 0
x(x – 2) –1(x – 2) = 0
(x – 2) (x – 1) = 0
x = 1 (or) 2 are the roots of the equation.

Question-14

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 x years.

His age 3 years ago = x – 3

5 years from now = x + 5

=

Taking LCM,

3(2x + 2) = (x – 3) (x + 5)

6x + 6 = x2 + 2x -15

x2 – 4x – 21 = 0

x2 – 7x + 3x - 21 = 0

x(x – 7) + 3(x – 7) = 0

x = 7 or x = -3

Age cannot be negative thus x = 7.

His present age = 7 years


Question-15

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 her marks in Maths = x and in English = y
Given x + y = 30 …………………. (1)
Given, (x + 2)(y – 3) = 210 …… (2)
From (1): x = 30 - y ……(3)
Substututing in (2) we get
(30 – y + 2) (y – 3) = 210
(32 – y) (y – 3) = 210
32y – 96 – y2 + 3y = 210
y2 – 35y + 306 = 0
y2 – 17y – 18y + 306 = 0
y(y – 17) – 18(y – 17) = 0
(y – 18) (y – 17) = 0
y = 18 (or) 17
 
x = 13 (when y= 17 by substituting in (3))
x = 12 (when y= 18 by substituting in (3))
Her marks in Maths is 13 and Science is 17 (or) Maths is 12 and Science is 18.

Question-16

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 x be the length of the shorter side
Longer side = (x + 30)
In the right angled Δ BCD, 
(BD) 2 = (BC)2 + (CD)2 (Pythogoras theorem)
(60 + x)2 = (x + 30)2 + x2
(x + 60)2 = x2 + x2 + 60x + 900
2x2 + 60x + 900 = x2 + 120x + 3600
x2 - 60x – 2700 = 0
x2 - 90 x + 30x – 2700 = 0
x(x - 90) + 30 (x - 90) = 0
(x + 30)(x – 90) = 0
x = -30; x = 90
Since the length of a side cannot be negative, x = 90
Shorter side = 90 m
Longer side = 90 + 30 = 120 m

Question-17

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 two numbers be x, y
x2 – y2 = 180 ………………. (1)
y2 = 8x ………………. (2)
Substitute (2) in (1)
x2 – 8x = 180
x2 – 8x – 180 = 0
x2 + 10x – 18x – 180 = 0
x(x + 10) – 18(x + 10) = 0
(x + 10)(x – 18) = 0
x = -10; x = 18
If x = -10 from (2)
y2 = 8x= 8(-10) = -80
This is not possible as square of any cannot be negative.
If x = 18 from (2)
y2 = 8x = 8(18) = 144
y = 12
Thus the numbers are 18, 12 (or) 18, -12.

Question-18

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 speed of the train = x km/hr

Distance traveled = 360 km

Time taken = =

If the speed increases by 5km/hr the time required to reach the place reduces to one hour

Time taken =

= 1

360

360

1800 = x2 + 5x

x2 + 5x – 1800 = 0

x2 + 45x – 40x - 1800 = 0

x(x + 45) – 40(x + 45) = 0

(x + 45)(x – 40) = 0

x = 40 (or) x = -45

But speed of the train can only be positive, thus x = 40 km/hr


Question-19

Two water taps together can fill a tank in 9 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 smaller tap = x hrs
In 1 hr of the tank can filled by smaller tap.

Let the time taken by larger tap = (x – 10) hrs
In 1 hr of the tank can be filled by larger tap.
& together they fill the tank in
=
=
75(2x - 10) = 8(x2 – 10x)
8x2 – 80x – 150x + 750 = 0
8x2 – 230x + 750 = 0
4x2 – 115x + 375 = 0 (Dividing by 2)
4x2 – 100x – 15x + 375 = 0
4x(x – 25) – 15(x – 25) = 0
4x – 15 = 0 (or) x – 25 = 0
x = (or) x = 25
x = cannot satisfy the condition
Small tap will fill the tank in 25 hrs and larger tap in 15 hrs.

Question-20

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 11 km/h more than that of the passenger train, find the average speed of the two trains.

Solution:
Let the average speed of the passenger train be x km/hr and the express train be (x + 11) km/hr
Time taken by passenger train =
Time taken by express train =
According to the question,
= 1
132
1452 = x2 + 11x
x2 +11x – 1452 = 0
x2 + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x – 33)(x + 44) = 0
x = 33, x = -44
Since the speed of train cannot be negative, x = 33 km/hr
Speed of the passenger train = 33 km/hr
Speed of the express train = 22km/hr

Question-21

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:
Let the side of the two squares be a, b.
Area of first square = a2
Area of second square = b2
a2 + b2 = 468 …………….. (1)
Perimeter of first square = 4a
Perimeter of second square = 4b
4a - 4b = 24
a – b = 6
a = b + 6 ……………….. (2)
substitute a in (1)
(b+6)2 + b2 = 468
b2 + 12b + 36 + b2 = 468
2b2 + 12b – 432 = 0
Dividing by 2,
b2 + 6b – 216 = 0
b2 + 18b – 12b – 216 = 0
b(b + 18) –12 (b +18) = 0
(b + 18)(b – 12) = 0
 
b = -18, b = 12
Side cannot be negative thus b = 12 m
Substitute b = 12 in (2)
a = b + 6 = 12 + 6 = 18 m
Side of first square = 18 m
Side of second square = 12 m

Question-22

Find nature of roots of the following quadratic equation. If real roots exists find them.
(i) 2x2 – 3x + 5 (ii) 3x2 - 4x + 4 (iii) 2x2 – 6x + 3 = 0
 

Solution:
(i) 2x2 – 3x + 5
Here a = 2, b = -3, c = 5
Discriminant Δ= b2 – 4ac = 9 - (4 × 2 × 5)
                     = 9 – 40
                     = -31
Equation 2x2 – 3x + 5 has imaginary roots.

(ii) 3x2 - 4x + 4
Here, a = 3, b = -4, c = 4
Δ = b2 – 4ac
   = (16 × 3) – (4 × 3 × 4)
   = (16 × 3) – (16 × 3)
   = 0
The roots are real and equal.
The roots are x =
                      =
                      =
                      =
                      =,
                      =,
                      = ,
 
(iii) 2x2 – 6x + 3 = 0
a = 2 , b = -6, c = 3
Discriminant Δ  = b2 – 4ac
                     = 36 – (4 × 2 × 3)
                     = 36 – 24
                     = 12
Δ > 0
Equation has real and distinct roots
x =
  =
  =
x =
x = (or) x =

Question-23

Find the value of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0  

Solution:
(i) 2x2 + kx + 3 = 0
As the equation has equal roots the value of Δ  = 0
a = 2 , b = k, c = 3
Δ = b2 – 4ac
0 = k2 – (4 × 2 × 3)
0 = k2 – 24
k2 = 24
k =
k = 2

(ii) kx2 – 2kx + 6 = 0
As the equation has equal roots the value of D = 0
a = k , b = -2k, c = 6
Δ = b2 – 4ac
0 = 4k2 – (4 × k × 6)
0 = 4k2 – 24k
0 = k2 – 6k
k(k – 6) = 0
k = 0 (or) k = 6
Thus k = 6.

Question-24

Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2 ? If so, find its length and breadth.

Solution:
Let length be x m and breadth be y m
Given, x = 2y .............(1)
ang xy = 800 m2..............(2)
substituting value of x from (1)
2y × y = 800
2y2 = 800
=> y2 – 400 = 0
y2 = 400
y = 20 m x = 40 m
Yes, it is possible to design a rectangular mango grove with length = 40 m and breadth = 20 m.

Question-25

Is the following situtation possible, If so determine the present ages. The sum of ages of 2 friends is 20 years. Four years ago, the product of their ages in year was 48.

Solution:
Let the present ages of friends be x, y.
x + y = 20 ................(1)
x = 20 – y
The product of their ages 4 yrs ago,
(x – 4)(y – 4) = 48
xy – 4y – 4x +16 = 48
xy – 4y – 4x = 32
(20 – y)y – 4y – 4(20 – y) = 32
20y – y2 – 4y – 80 + 4y = 32
-y2 + 20y – 112 = 0
Δ = b2 – 4ac
   = 400 – 4 × 1 × -112
   = 400 – 448
   = -48
As Δ < 0 is not possible.

Question-26

Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so find its length and breadth.  

Solution:
Area = Length × Breadth
       = l × b = 400 m2...................(1)
Perimeter = 2(l + b) = 80 m
2(l + b) = 80
l + b = 40 ...................(2)
From (1), b =
substituting value of b in equation (2) 

l + = 40
l2 + 400 = 40l
l2 – 40l + 400 = 0
Δ = b2 – 4ac
   = (-40)2 – (4 × 400 × 1)
   = 1600 – 1600
   = 0
Equation has equal roots.
As the equation has equal roots we can only design a square plot.




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