CBSE NCERT Solutions for Class 10 Mathematics Chapter 4β¬Β¦Β Β· Thus, the given equation is a...
Transcript of CBSE NCERT Solutions for Class 10 Mathematics Chapter 4β¬Β¦Β Β· Thus, the given equation is a...
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CBSE NCERT Solutions for Class 10 Mathematics Chapter 4 Back of Chapter Questions
1. Check whether the following are quadratic equations:
(i) (π₯π₯ + 1)2 = 2(π₯π₯ β 3)
(ii) π₯π₯2 β 2π₯π₯ = (β2)(3β π₯π₯)
(iii) (π₯π₯ β 2)(π₯π₯ + 1) = (π₯π₯ β 1)(π₯π₯ + 3)
(iv) (π₯π₯ β 3)(2π₯π₯ + 1) = π₯π₯(π₯π₯ + 5)
(v) (2π₯π₯ β 1)(π₯π₯ β 3) = (π₯π₯ + 5)(π₯π₯ β 1)
(vi) π₯π₯2 + 3π₯π₯ + 1 = (π₯π₯ β 2)2
(vii) (π₯π₯ + 2)3 = 2π₯π₯(π₯π₯2 β 1)
(viii) π₯π₯3 β 4π₯π₯2 β π₯π₯ + 1 = (π₯π₯ β 2)3
Solution:
(i) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called aquadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: (π₯π₯ + 1)2 = 2(π₯π₯ β 3)
Using the formula (ππ + ππ)2 = ππ2 + 2ππππ + ππ2
β π₯π₯2 + 2π₯π₯ + 1 = 2π₯π₯ β 6
β π₯π₯2 + 7 = 0
Here, ππ = 1, ππ = 0 and ππ = 7.
Thus, the given equation is a quadratic equation as ππ β 0.
(ii) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called aquadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: π₯π₯2 β 2π₯π₯ = (β2)(3β π₯π₯)
β π₯π₯2 β 2π₯π₯ = β6 + 2π₯π₯
β π₯π₯2 β 4π₯π₯ + 6 = 0
Here, ππ = 1, ππ = β4 and ππ = 6.
Thus, the given equation is a quadratic equation as ππ β 0.
(iii) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called aquadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
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Given equation: (π₯π₯ β 2)(π₯π₯ + 1) = (π₯π₯ β 1)(π₯π₯ + 3)
β π₯π₯2 β π₯π₯ β 2 = π₯π₯2 + 2π₯π₯ β 3
β 3π₯π₯ β 1 = 0
But, here ππ = 0.
So, the given equation is not a quadratic equation.
(iv) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called a quadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: (π₯π₯ β 3)(2π₯π₯ + 1) = π₯π₯(π₯π₯ + 5)
β 2π₯π₯2 β 5π₯π₯ β 3 = π₯π₯2 + 5π₯π₯
β π₯π₯2 β 10π₯π₯ β 3 = 0
Here, ππ = 1, ππ = β10 and ππ = β3.
Thus, the given equation is a quadratic equation as ππ β 0.
(v) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called a quadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: (2π₯π₯ β 1)(π₯π₯ β 3) = (π₯π₯ + 5)(π₯π₯ β 1)
β 2π₯π₯2 β 7π₯π₯ + 3 = π₯π₯2 + 4π₯π₯ β 5
β π₯π₯2 β 11π₯π₯ + 8 = 0
Here, ππ = 1, ππ = β11 and ππ = 8.
Thus, the given equation is a quadratic equation as ππ β 0.
(vi) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called a quadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: π₯π₯2 + 3π₯π₯ + 1 = (π₯π₯ β 2)2
Using the formula (ππ β ππ)2 = ππ2 β 2ππππ + ππ2
β π₯π₯2 + 3π₯π₯ + 1 = π₯π₯2 β 4π₯π₯ + 4
β 7π₯π₯ β 3 = 0
But, here ππ = 0.
So, the given equation is not a quadratic equation.
(vii) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called a quadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: (π₯π₯ + 2)3 = 2π₯π₯(π₯π₯2 β 1)
Using the formula (ππ + ππ)3 = ππ3 + ππ3 + 3ππ2ππ + 3ππππ2
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β π₯π₯3 + 8 + 6π₯π₯2 + 12π₯π₯ = 2π₯π₯3 β 2π₯π₯
β π₯π₯3 β 14π₯π₯ β 6π₯π₯2 β 8 = 0
This equation is not of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0
So, the given equation is not a quadratic equation.
(viii) We know that any equation of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0 is called a quadratic equation, where ππ, ππ, ππ are real numbers and ππ β 0.
Given equation: π₯π₯3 β 4π₯π₯2 β π₯π₯ + 1 = (π₯π₯ β 2)3
Using the formula (ππ β ππ)3 = ππ3 β 3ππ2ππ + 3ππππ2 β ππ3
β π₯π₯3 β 4π₯π₯2 β π₯π₯ + 1 = π₯π₯3 β 8 β 6π₯π₯2 + 12π₯π₯
β 2π₯π₯2 β 13π₯π₯ + 9 = 0
Here, ππ = 2, ππ = β13 and ππ = 9.
Thus, the given equation is a quadratic equation as ππ β 0.
2. 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.
(iii) Rohanβs 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 Rohanβs present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km hβ 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 be π₯π₯ m.
Hence, the length of the plot is (2π₯π₯ + 1) m. (Since, given that length is one more than twice its breadth)
Therefore, area of a rectangle = length Γ breadth
Given: area of rectangle = 528 m2
β΄ 528 = π₯π₯(2π₯π₯ + 1)
β 2π₯π₯2 + π₯π₯ β 528 = 0 β¦β¦β¦(i), which is of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0
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Here ππ = 2(β 0), ππ = 1 and ππ = β528
Thus, quadratic equation (i) represents the situation given in the question and roots of this equation will represent the breadth of the plot.
(ii) We know that the difference between two consecutive positive integers is 1.
So, let the consecutive positive integers be π₯π₯ and π₯π₯ + 1.
Given that their product is 306.
β΄ π₯π₯(π₯π₯ + 1) = 306
β π₯π₯2 + π₯π₯ β 306 = 0 β¦β¦β¦.. (i), which is of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0
Here ππ = 1(β 0),ππ = 1 and ππ = β306
Thus, quadratic equation (i) represents the situation given in the question and roots of this equation will represent the smaller positive integer.
(iii) Let Rohan's age be π₯π₯,
His mother's age = π₯π₯ + 26 (given that Rohanβs mother is 26 years older than him)
3 years from now:
Rohan's age will be = π₯π₯ + 3
Mother's age will be = π₯π₯ + 26 + 3 = π₯π₯ + 29
Also given that the product of their ages after 3 years is 360.
β΄ (π₯π₯ + 3)(π₯π₯ + 29) = 360
On simplification, we get
π₯π₯2 + 32π₯π₯ β 273 = 0β¦β¦β¦ (i), which is of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0
Here ππ = 1(β 0),ππ = 32 and ππ = β273
Thus, quadratic equation (i) represents the situation given in the question and positive root of this equation will represent the Rohanβs present age.
(iv) In first case,
Let the speed of train be π₯π₯ km/h.
Total time taken to travel 480 km = 480π₯π₯
hrs
In second case,
Given: speed became 8 km/h less
So, the speed of train = (π₯π₯ β 8)km/h
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Also given that the train will take 3 more hours to cover the same distance.
Therefore, time take to travel 480 km = οΏ½480π₯π₯
+ 3οΏ½hrs
Speed Γ Time = Distance
(π₯π₯ β 8) οΏ½480π₯π₯ + 3οΏ½ = 480
β 480 + 3π₯π₯ β3840π₯π₯ β 24 = 480
β 3π₯π₯ β3840π₯π₯ = 24
β 3π₯π₯2 β 24π₯π₯ β 3840 = 0
β π₯π₯2 β 8π₯π₯ β 1280 = 0β¦β¦.. (i), which is of the form πππ₯π₯2 + πππ₯π₯ + ππ = 0
Here ππ = 1(β 0),ππ = β8 and ππ = β1280
Thus, quadratic equation (i) represents the situation given in the question and positive root of this equation will represent the speed of train.
EXERCISE 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) π₯π₯2 β 3π₯π₯ β 10 = 0
(ii) 2π₯π₯2 + π₯π₯ β 6 = 0
(iii) β2 π₯π₯2 + 7π₯π₯ + 5β2 = 0
(iv) 2π₯π₯2 β π₯π₯ + 18
= 0
(v) 100π₯π₯2 β 20π₯π₯ + 1 = 0
Solution:
(i) To find the roots of given quadratic equation, lets first factorise the given quadratic expression π₯π₯2 β 3π₯π₯ β 10. The given quadratic expression can be written as follows:
π₯π₯2 β 3π₯π₯ β 10
= π₯π₯2 β 5π₯π₯ + 2π₯π₯ β 10 (we factorise by method of splitting the middle term)
= π₯π₯(π₯π₯ β 5) + 2(π₯π₯ β 5)
= (π₯π₯ β 5)(π₯π₯ + 2)
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Now, the roots of this quadratic equation are the values of π₯π₯ for which (π₯π₯ β 5)(π₯π₯ + 2) = 0
β΄ π₯π₯ β 5 = 0 or π₯π₯ + 2 = 0
ππ. ππ. ,π₯π₯ = 5 or π₯π₯ = β2
Hence, the roots of this quadratic equation are 5 andβ2.
(ii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2π₯π₯2 + π₯π₯ β 6 . The given quadratic expression can be written as follows:
2π₯π₯2 + π₯π₯ β 6
= 2π₯π₯2 + 4π₯π₯ β 3π₯π₯ β 6 (we factorise by method of splitting the middle term)
= 2π₯π₯(π₯π₯ + 2)β 3(π₯π₯ + 2)
= (π₯π₯ + 2)(2π₯π₯ β 3)
Now, the roots of this quadratic equation are the values of π₯π₯ for which (π₯π₯ + 2)(2π₯π₯ β 3) = 0
β΄ π₯π₯ + 2 = 0 or 2π₯π₯ β 3 = 0
ππ. ππ. ,π₯π₯ = β2 or π₯π₯ = 32
Hence, the roots of this quadratic equation are β2 and 32.
(iii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression β2π₯π₯2 + 7π₯π₯ + 5β2 . The given quadratic expression can be written as follows:
β2π₯π₯2 + 7π₯π₯ + 5β2
= β2π₯π₯2 + 5π₯π₯ + 2π₯π₯ + 5β2 (we factorise by method of splitting the middle term)
= π₯π₯οΏ½β2π₯π₯ + 5οΏ½ + β2οΏ½β2π₯π₯ + 5οΏ½
= οΏ½β2π₯π₯ + 5οΏ½οΏ½π₯π₯ + β2οΏ½
Now, the roots of this quadratic equation are the values of π₯π₯ for which
(β2π₯π₯ + 5)οΏ½π₯π₯ + β2οΏ½ = 0
β΄ β2π₯π₯ + 5 = 0 or π₯π₯ + β2 = 0
ππ. ππ. ,π₯π₯ = β 5β2
or π₯π₯ = ββ2
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Hence, the roots of this quadratic equation are β 5β2
and ββ2.
(iv) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2π₯π₯2 β π₯π₯ + 1
8 . The given quadratic expression can be
written as follows:
2π₯π₯2 β π₯π₯ +18
=18
(16π₯π₯2 β 8π₯π₯ + 1)
= 18
(16π₯π₯2 β 4π₯π₯ β 4π₯π₯ + 1) (we factorise by method of splitting the middle term)
=18οΏ½4π₯π₯(4π₯π₯ β 1)β 1(4π₯π₯ β 1)οΏ½
=18
(4π₯π₯ β 1)2
Now, the roots of this quadratic equation are the values of π₯π₯ for which (4π₯π₯ β 1)2 = 0
Thus, (4π₯π₯ β 1) = 0 or (4π₯π₯ β 1) = 0
ππ. ππ. ,π₯π₯ = 14 or π₯π₯ = 1
4
Hence, the roots of this quadratic equation are 14 and 1
4.
(v) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 100π₯π₯2 β 20π₯π₯ + 1. The given quadratic expression can be written as follows:
100π₯π₯2 β 20π₯π₯ + 1
= 100π₯π₯2 β 10π₯π₯ β 10π₯π₯ + 1 (we factorise by method of splitting the middle term)
= 10π₯π₯(10π₯π₯ β 1)β 1(10π₯π₯ β 1)
= (10π₯π₯ β 1)2
Now, the roots of this quadratic equation are the values of π₯π₯ for which (10π₯π₯ β 1)2 = 0
Thus, (10π₯π₯ β 1) = 0 or (10π₯π₯ β 1) = 0
i. e., π₯π₯ = 110
or π₯π₯ = 110
Hence, the roots of this quadratic equation are 110
and 110
.
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2. Solve the problems given below.
Represent the following situations 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 βΉ 750. We would like to find out the number of toys produced on that day.
Solution:
(i) Let the number of John's marbles be π₯π₯.
So, the number of Jivanti's marbles = 45β π₯π₯
If both lost 5 marbles each,
Then number of marbles left with John = π₯π₯ β 5
Then number of marbles left with Jivanti = 45β π₯π₯ β 5 = 40 β π₯π₯
Given that the product of their marbles is 124.
β΄ (π₯π₯ β 5)(40β π₯π₯) = 124
β π₯π₯2 β 45π₯π₯ + 324 = 0
β π₯π₯2 β 36π₯π₯ β 9π₯π₯ + 324 = 0
β π₯π₯(π₯π₯ β 36)β 9(π₯π₯ β 36) = 0
β (π₯π₯ β 36)(π₯π₯ β 9) = 0
Either π₯π₯ = 36 = 0 or π₯π₯ β 9 = 0
ππ. ππ. ,π₯π₯ = 36 or π₯π₯ = 9
If the number of John's marbles = 36
Then, the number of Jivanti's marbles = 45β 36 = 9
If the number of John's marbles = 9
Then, the number of Jivanti's marbles = 45β 9 = 36.
(ii) Let the number of toys produced on that day be π₯π₯.
β΄ The cost of production of each toy that day = βΉ (55β π₯π₯)
So, the total cost of production that day = π₯π₯(55β π₯π₯)
As per the question, the total cost of production of the toys = βΉ 750
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β΄ (55β π₯π₯)π₯π₯ = 750
β π₯π₯2 β 55π₯π₯ + 750 = 0
β π₯π₯2 β 25π₯π₯ β 30π₯π₯ + 750 = 0
β π₯π₯(π₯π₯ β 25)β 30(π₯π₯ β 25) = 0
β (π₯π₯ β 25)(π₯π₯ β 30) = 0
Either π₯π₯ β 25 = 0 or π₯π₯ β 30 = 0
ππ. ππ. ,π₯π₯ = 25 or π₯π₯ = 30
Thus, the number of toys produced that day will be either 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first number be π₯π₯.
Then the second number is 27 β π₯π₯. (Given sum of two numbers = 27)
Thus, their product = π₯π₯(27β π₯π₯)
According to the question, the product of these numbers is 182.
Therefore, π₯π₯(27β π₯π₯) = 182
β π₯π₯2 β 27π₯π₯ + 182 = 0
β π₯π₯2 β 13π₯π₯ β 14π₯π₯ + 182 = 0
β π₯π₯(π₯π₯ β 13)β 14(π₯π₯ β 13) = 0
β (π₯π₯ β 13)(π₯π₯ β 14) = 0
Either π₯π₯ = 13 = 0 or π₯π₯ β 14 = 0
i. e. , π₯π₯ = 13 or π₯π₯ = 14
If first number = 13, then
Second number = 27β 13 = 14
If first number = 14, then
Second number = 27 β 14 = 13
Hence, the numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Solution:
We know that the difference between two consecutive positive integers is 1.
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So, let the consecutive positive integers be π₯π₯ and π₯π₯ + 1.
As per the question, π₯π₯2 + (π₯π₯ + 1)2 = 365
β π₯π₯2 + π₯π₯2 + 1 + 2π₯π₯ = 365
β 2π₯π₯2 + 2π₯π₯ β 364 = 0
β π₯π₯2 + π₯π₯ β 182 = 0
β π₯π₯2 + 14π₯π₯ β 13π₯π₯ β 182 = 0
β π₯π₯(π₯π₯ + 14)β 13(π₯π₯ + 14) = 0
β (π₯π₯ + 14)(π₯π₯ β 13) = 0
Either π₯π₯ + 14 = 0 or π₯π₯ β 13 = 0, i. e. , π₯π₯ = β14 or π₯π₯ = 13
Since given that integers are positive, π₯π₯ can only be 13.
β΄ π₯π₯ + 1 = 13 + 1 = 14
Hence, the two consecutive positive integers are 13 and 14.
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: As per the question, hypotenuse is 13 cm
Let the base of the right triangle be π₯π₯ cm.
Its altitude = (π₯π₯ β 7)cm
From Pythagoras theorem,
Base2 + Altitude2 = Hypotenuse2
β΄ π₯π₯2 + (π₯π₯ β 7)2 = 132
β π₯π₯2 + π₯π₯2 + 49 β 14π₯π₯ = 169
β 2π₯π₯2 β 14π₯π₯ β 120 = 0
β π₯π₯2 β 7π₯π₯ β 60 = 0
β π₯π₯2 β 12π₯π₯ + 5π₯π₯ β 60 = 0
β π₯π₯(π₯π₯ β 12) + 5(π₯π₯ β 12) = 0
β (π₯π₯ β 12)(π₯π₯ + 5) = 0
Either π₯π₯ β 12 = 0 or π₯π₯ + 5 = 0, i. e. , π₯π₯ = 12 or π₯π₯ = β5
Since sides of a triangle are positive, π₯π₯ can only take 12.
Hence, the base of the right triangle is 12 cm and the altitude of this triangle is (12β 7)cm = 5 cm.
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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 βΉ. 90, find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced on that day be π₯π₯.
So, the cost of production of each article = βΉ(2π₯π₯ + 3)
According to the question, the total cost of production on that day was βΉ 90.
We know that
Total cost of production = Cost of each article Γ Number of articles produced
β΄ π₯π₯(2π₯π₯ + 3) = 90
β 2π₯π₯2 + 3π₯π₯ β 90 = 0
β 2π₯π₯2 + 15π₯π₯ β 12π₯π₯ β 90 = 0
β π₯π₯(2π₯π₯ + 15) β 6(2π₯π₯ + 15) = 0
β (2π₯π₯ + 15)(π₯π₯ β 6) = 0
Either 2π₯π₯ + 15 = 0 or π₯π₯ β 6 = 0, ππ. ππ. ,π₯π₯ = β152
or π₯π₯ = 6
Itβs clear that number of articles produced can only be a positive integer, so, π₯π₯ can only be 6.
Therefore, number of articles produced on that day = 6
Cost of each article = (2 Γ 6) + 3 = βΉ 15
EXERCISE 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2π₯π₯2 β 7π₯π₯ + 3 = 0
(ii) (ii) 2π₯π₯2 + π₯π₯ β 4 = 0
(iii) 4π₯π₯2 + 4β3π₯π₯ + 3 = 0
(iv) 2π₯π₯2 + π₯π₯ + 4 = 0
Solution:
(i) Given quadratic equation: 2π₯π₯2 β 7π₯π₯ + 3 = 0
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β 2π₯π₯2 β 7π₯π₯ = β3
On dividing both sides of the equation by 2, we obtain
β π₯π₯2 β72 π₯π₯ = β
32
β π₯π₯2 β 2 Γ π₯π₯ Γ74 = β
32
On adding οΏ½74οΏ½2 to both sides of equation by, we obtain
β (π₯π₯)2 β 2 Γ π₯π₯ Γ74 + οΏ½
74οΏ½
2
= οΏ½74οΏ½
2
β32
β οΏ½π₯π₯ β74οΏ½
2
=4916β
32
β οΏ½π₯π₯ β74οΏ½
2
=2516
β οΏ½π₯π₯ β 74οΏ½ = Β± 5
4 (Cancelling square both the sides)
β π₯π₯ =74 Β±
54
β π₯π₯ = 74
+ 54 or π₯π₯ = 7
4β 5
4
β π₯π₯ = 124
or π₯π₯ = 24
β π₯π₯ = 3 or 12
Hence, the roots of this quadratic equation are 3 and 12 .
(ii) 2π₯π₯2 + π₯π₯ β 4 = 0
β 2π₯π₯2 + π₯π₯ = 4
On dividing both sides of the equation by 2, we obtain
β π₯π₯2 +12 π₯π₯ = 2
On adding οΏ½14οΏ½2 to both sides of the equation, we obtain
β (π₯π₯)2 + 2 Γ π₯π₯ Γ14 + οΏ½
14οΏ½
2
= 2 + οΏ½14οΏ½
2
β οΏ½π₯π₯ +14οΏ½
2
=3316
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β π₯π₯ + 14
= Β± β334
(Cancelling square both the sides)
β π₯π₯ = Β±β33
4 β14
β π₯π₯ =Β±β33β 1
4
β π₯π₯ = β33β14
or ββ33β14
Hence, the roots of this quadratic equation are β1+β334
and β1ββ334
.
(iii) 4π₯π₯2 + 4β3π₯π₯ + 3 = 0
β (2π₯π₯)2 + 2 Γ 2π₯π₯ Γ β3 + οΏ½β3οΏ½2
= 0
β οΏ½2π₯π₯ + β3οΏ½2
= 0
β οΏ½2π₯π₯ + β3οΏ½
= 0 and οΏ½2π₯π₯ + β3οΏ½
= 0
β π₯π₯ = ββ32
and π₯π₯ = ββ32
Hence, the roots of this quadratic equation are ββ32
and β β32
.
(iv) 2π₯π₯2 + π₯π₯ + 4 = 0
β 2π₯π₯2 + π₯π₯ = β4
On dividing both sides of the equation by 2, we obtain
β π₯π₯2 +12 π₯π₯ = β2
β π₯π₯2 + 2 Γ π₯π₯ Γ14 = β2
On adding οΏ½14οΏ½2to both sides of the equation, we obtain
β (π₯π₯)2 + 2 Γ π₯π₯ Γ14 + οΏ½
14οΏ½
2
= οΏ½14οΏ½
2
β 2
β οΏ½π₯π₯ +14οΏ½
2
=1
16β 2
β οΏ½π₯π₯ +14οΏ½
2
= β3116
Since, the square of a number cannot be negative.
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Thus, there is no real root for the given equation.
2. Find the roots of the quadratic equations given by applying the quadratic formula.
(i) 2π₯π₯2 β 7π₯π₯ + 3 = 0
(ii) 2π₯π₯2 + π₯π₯ β 4 = 0
(iii) 4π₯π₯2 + 4β3π₯π₯ + 3 = 0
(iv) 2π₯π₯2 + π₯π₯ + 4 = 0
Solution:
(i) 2π₯π₯2 β 7π₯π₯ + 3 = 0
On comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 2,ππ = β7, ππ = 3
By using quadratic formula, we obtain
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
β π₯π₯ =7 Β± β49β 24
2ππ
β π₯π₯ =7 Β± β25
4
β π₯π₯ =7 Β± 5
4
β π₯π₯ = 7+54
or 7β54
β π₯π₯ = 124
or 24
β΄ π₯π₯ = 3 or 12
Hence, the roots of this quadratic equation are 3 and 12 .
(ii) 2π₯π₯2 + π₯π₯ β 4 = 0
On comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 2,ππ = 1, ππ = β4
By using quadratic formula, we obtain
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
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β π₯π₯ =β1 Β± β1 + 32
4
β π₯π₯ =β1 Β± β33
4
β΄ π₯π₯ = β1+β334
or β1ββ334
Hence, the roots of this quadratic equation are β1+β334
and β1ββ334
.
(iii) 4π₯π₯2 + 4β3π₯π₯ + 3 = 0
On comparing this equation πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain ππ = 4,ππ =4β3, ππ = 3
By using quadratic formula, we obtain
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
β π₯π₯ =β4β3 Β± β48β 48
8
β π₯π₯ =β4β3 Β± 0
8
β΄ π₯π₯ = ββ32
or β β32
Hence, the roots of this quadratic equation are ββ32
and β β32
.
(iv) 2π₯π₯2 + π₯π₯ + 4 = 0
On comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 2,ππ = 1, ππ = 4
By using quadratic formula, we obtain
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
β π₯π₯ =β1 Β± β1 β 32
4
β π₯π₯ =β1 Β± ββ31
4
Hence, roots do not exist for this quadratic equation as π·π· = ππ2 β 4ππππ = β31 < 0.
3. Find the roots of the following equations:
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(i) π₯π₯ β 1π₯π₯
= 3, π₯π₯ β 0
(ii) 1π₯π₯+4
β 1π₯π₯β7
= 1130
, π₯π₯ β β4, 7
Solution:
(i) π₯π₯ β 1π₯π₯
= 3 β π₯π₯2 β 3π₯π₯ β 1 = 0
On comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 1,ππ = β3, ππ = β1
By using quadratic formula, we obtain
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
β π₯π₯ =3 Β± β9 + 4
2
β π₯π₯ =3 Β± β13
2
Therefore, π₯π₯ = 3+β132
or 3ββ132
(ii) 1π₯π₯+4
β 1π₯π₯β7
= 1130
βπ₯π₯ β 7β π₯π₯ β 4
(π₯π₯ + 4)(π₯π₯ β 7) =1130
ββ11
(π₯π₯ + 4)(π₯π₯ β 7) =1130
β (π₯π₯ + 4)(π₯π₯ β 7) = β30
β π₯π₯2 β 3π₯π₯ β 28 = β30
β π₯π₯2 β 3π₯π₯ + 2 = 0
β π₯π₯2 β 2π₯π₯ β π₯π₯ + 2 = 0
β π₯π₯(π₯π₯ β 2)β 1(π₯π₯ β 2) = 0
β (π₯π₯ β 2)(π₯π₯ β 1) = 0
β π₯π₯ = 1 or 2
4. The sum of the reciprocals of Rehmanβs ages, (in years) 3 years ago and 5 years from now is 1
3. Find his present age.
Solution:
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Let Rehmanβs present age = π₯π₯ years.
His age three years ago = (π₯π₯ β 3) years.
His age five years from now = (π₯π₯ + 5) years.
As per the question, the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1
3.
β΄1
π₯π₯ β 3 +1
π₯π₯ + 5 =13
π₯π₯ + 5 + π₯π₯ β 3(π₯π₯ β 3)(π₯π₯ + 5) =
13
2π₯π₯ + 2(π₯π₯ β 3)(π₯π₯ + 5) =
13
β 3(2π₯π₯ + 2) = (π₯π₯ β 3)(π₯π₯ + 5)
β 6π₯π₯ + 6 = π₯π₯2 + 2π₯π₯ β 15
β π₯π₯2 β 4π₯π₯ β 21 = 0
β π₯π₯2 β 7π₯π₯ + 3π₯π₯ β 21 = 0
β π₯π₯(π₯π₯ β 7) + 3(π₯π₯ β 7) = 0
β (π₯π₯ β 7)( π₯π₯ + 3) = 0
β π₯π₯ = 7 or π₯π₯ = β3
Itβs clear that age is always positive.
Thus, Rehman's present age is 7 years.
5. In a class test, the sum of Shefaliβs 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 Shefaliβs marks in Mathematics = π₯π₯.
Then, her marks in English = 30β π₯π₯. (Given in question)
As per the question, we get
(π₯π₯ + 2)(30β π₯π₯ β 3) = 210
(π₯π₯ + 2)(27β π₯π₯) = 210
β βπ₯π₯2 + 25π₯π₯ + 54 = 210
β π₯π₯2 β 25π₯π₯ + 156 = 0
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β π₯π₯2 β 12π₯π₯ β 13π₯π₯ + 156 = 0
β π₯π₯(π₯π₯ + 2)β 13(π₯π₯ β 12) = 0
β (π₯π₯ β 12)(π₯π₯ β 13) = 0
β π₯π₯ = 12 or π₯π₯ = 13
If the marks in Mathematics is 12, then marks in English will be 30β 12 = 18
If the marks in Mathematics is 13,then marks in English will be 30β 13 = 17
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangle be π₯π₯ m.
Then, longer side of the rectangle = (π₯π₯ + 30)m
Diagonal of the rectangle = οΏ½π₯π₯2 + (π₯π₯ + 30)2 (By Pythagoras theorem)
But in question, it is given that the diagonal of the rectangular field is 60 ππ more than the shorter side.
β΄ οΏ½π₯π₯2 + (π₯π₯ + 30)2 = π₯π₯ + 60
β π₯π₯2 + (π₯π₯ + 30)2 = (π₯π₯ + 60)2 (By squaring on both the sides)
β π₯π₯2 + π₯π₯2 + 900 + 60π₯π₯ = π₯π₯2 + 3600 + 120π₯π₯
β π₯π₯2 β 60π₯π₯ β 2700 = 0
β π₯π₯2 β 90π₯π₯ + 30π₯π₯ β 2700 = 0
β π₯π₯(π₯π₯ β 90) + 30(π₯π₯ β 90)
β (π₯π₯ β 90)(π₯π₯ + 30) = 0
β π₯π₯ = 90 or π₯π₯ = β30
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But, side of a rectangle cannot be negative. So, the length of the shorter side is 90 m. Thus, the length of the longer side will be (90 + 30) m = 120 m
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 and smaller number be π₯π₯ and π¦π¦ respectively.
It is given in the question that,
π₯π₯2 β π¦π¦2 = 180 and π¦π¦2 = 8π₯π₯
β π₯π₯2 β 8π₯π₯ = 180
β π₯π₯2 β 8π₯π₯ β 180 = 0
β π₯π₯2 β 18π₯π₯ + 10π₯π₯ β 180 = 0
β π₯π₯(π₯π₯ β 18) + 10(π₯π₯ β 18) = 0
β (π₯π₯ β 18)(π₯π₯ + 10) = 0
β π₯π₯ = 18,β10
If larger number, π₯π₯ = β10
then smaller number, π¦π¦ = Β±β8π₯π₯
= Β±οΏ½8(β10)
= Β±ββ80
Since we cannot have negative number in roots
π₯π₯ = β10 is not possible
Therefore, the larger number will be 18 only.
π₯π₯ = 18
β΄ π¦π¦2 = 8π₯π₯ = 8 Γ 18 = 144
β π¦π¦ = Β±β144 = Β±12
β΄ Smaller number = Β±12
Therefore, the numbers are 18 and 12 or 18 and β12.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km hβ more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the speed to the train be π₯π₯ km/h.
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Total time taken to cover 360 km = 360π₯π₯
hr
It is given in question,
(π₯π₯ + 5) οΏ½360π₯π₯β 1οΏ½ = 360 (Distance = Speed Γ Time)
β 360β π₯π₯ +1800π₯π₯ β 5 = 360
β π₯π₯2 + 5π₯π₯ β 1800 = 0
β π₯π₯2 + 45π₯π₯ β 40π₯π₯ β 1800 = 0
β π₯π₯(π₯π₯ + 45)β 40(π₯π₯ + 45) = 0
β (π₯π₯ + 45)(π₯π₯ β 40) = 0
β π₯π₯ = 40 or π₯π₯ = β45
But, speed cannot be negative.
Thus, the speed of the train is 40 km/h
9. Two water taps together can fill a tank in 9 38 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 the smaller diameter tap to fill the tank be π₯π₯ hr.
Time taken by the larger diameter tap = (π₯π₯ β 10)hr
Volume of tank filled by smaller tap in 1hour = 1π₯π₯
Volume of tank filled by larger tap in 1 hour = 1π₯π₯β10
As per the question, the tank can be filled in 9 38
= 758
hours by both the taps together.
Hence,
1π₯π₯ +
1π₯π₯ β 10 =
875
π₯π₯ β 10 + π₯π₯π₯π₯(π₯π₯ β 10) =
875
β2π₯π₯ β 10π₯π₯(π₯π₯ β 10) =
875
β 75(2π₯π₯ β 10) = 8π₯π₯2 β 80π₯π₯
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β 150π₯π₯ β 750 = 8π₯π₯2 β 80π₯π₯
β 8π₯π₯2 β 230π₯π₯ + 750 = 0
β 8π₯π₯2 β 200π₯π₯ β 30π₯π₯ + 750 = 0
β 8π₯π₯(π₯π₯ β 25) β 30(π₯π₯ β 25) = 0
β (π₯π₯ β 25)(8π₯π₯ β 30) = 0
ππ. ππ. ,π₯π₯ = 25 or π₯π₯ = 308
= 154
Taking π₯π₯ = 154
Time taken by smaller tap
= π₯π₯ = 154
hrs
Time taken by larger tap = π₯π₯ β 10
= 154β 10 = 15β40
4= β 25
4
Since time is negative,
π₯π₯ = 154
is not the solution
Thus, the time taken separately by the smaller diameter tap and the larger diameter tap will be 25 hours and (25β 10) = 15 hours respectively.
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 11 km hβ more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of passenger train be π₯π₯ km/h.
Average speed of express train = (π₯π₯ + 11) km/h (Given in question)
According to the question, the time taken by the express train to cover 132 km is
1 hour less than a passenger train.
Therefore, time taken by passenger train β time taken by express train = 1 hour
β΄ 132π₯π₯β 132
π₯π₯+11= 1 οΏ½Total time = Distance
Average SpeedοΏ½
β 132 οΏ½π₯π₯ + 11 β π₯π₯π₯π₯(π₯π₯ + 11) οΏ½ = 1
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β132 Γ 11π₯π₯(π₯π₯ + 11) = 1
β 132 Γ 11 = π₯π₯(π₯π₯ + 11)
β π₯π₯2 + 11π₯π₯ β 1452 = 0
β π₯π₯2 + 44π₯π₯ β 33π₯π₯ β 1452 = 0
β π₯π₯(π₯π₯ + 44)β 339π₯π₯ + 44) = 0
β (π₯π₯ + 44)(π₯π₯ β 33) = 0
β π₯π₯ = β44, 33
Average speed of passenger train cannot be negative. Hence, the speed of the passenger train is 33km/h and thus, the speed of the express train will be 33 +11 = 44 km/h.
11. 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 sides of the two squares be π₯π₯ m and π¦π¦ m.
So, their perimeter will be 4π₯π₯ and 4π¦π¦ respectively and their areas will be π₯π₯2and π¦π¦2 respectively.
According to the question, 4π₯π₯ β 4π¦π¦ = 24
β π₯π₯ β π¦π¦ = 6
β π₯π₯ = π¦π¦ + 6
Also, π₯π₯2 + π¦π¦2 = 468
β (6 + π¦π¦)2 + π¦π¦2 = 468
β 36 + π¦π¦2 + 12π¦π¦ + π¦π¦2 = 468
β 2π¦π¦2 + 12π¦π¦ β 432 = 0
β π¦π¦2 + 6π¦π¦ β 216 = 0
β π¦π¦2 + 18π¦π¦ β 12π¦π¦ β 216 = 0
β π¦π¦(π¦π¦ + 18)β 12(π¦π¦ + 18) = 0
β (π¦π¦ + 18)(π¦π¦ β 12) = 0
β π¦π¦ = β18 or 12.
But, side of a square cannot be negative.
Hence, π¦π¦ = 12 & π₯π₯ = 12 + 6 = 18
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Therefore, the sides of the squares are 12 m and 18 m
EXERCISE 4.4
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2π₯π₯2 β 3π₯π₯ + 5 = 0
(ii) 3π₯π₯2 β 4β3π₯π₯ + 4 = 0
(iii) 2π₯π₯2 β 6π₯π₯ + 3 = 0
Solution:
We know that for a quadratic equation πππ₯π₯2 + πππ₯π₯ + ππ = 0, discriminant is
ππ2 β 4ππππ.
(A) If ππ2 β 4ππππ > 0 implies two distinct real roots
(B) If ππ2 β 4ππππ = 0 implies two equal real roots
(C) If ππ2 β 4ππππ < 0 implies imaginary roots
(i) 2π₯π₯2 β 3π₯π₯ + 5 = 0
Comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain ππ = 2, ππ =β3, ππ = 5
Discriminant = ππ2 β 4ππππ = (β3)2 β 4(2)(5) = 9β 40 = β31
As ππ2 β 4ππππ < 0,
Hence, no real root is possible for the given equation.
(ii) 3π₯π₯2 β 4β3π₯π₯ + 4 = 0
Comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 3,ππ = β4β3, ππ = 4
Discriminant = ππ2 β 4ππππ = οΏ½β4β3οΏ½2β 4(3)(4) = 48β 48 = 0
As ππ2 β 4ππππ = 0,
So, real roots exist for the given equation and they are equal to each other and the roots will be β ππ
2ππ and β ππ
2ππ.
βππ
2ππ =βοΏ½β4β3οΏ½
2 Γ 3 =4β3
6 =2β3
3 =2β3
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Hence, the roots are 2β3
and 2β3
.
(iii) 2π₯π₯2 β 6π₯π₯ + 3 = 0
Comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain ππ = 2, ππ =β6, ππ = 3
Discriminant = ππ2 β 4ππππ = (β6)2 β 4(2)(3) = 36β 24 = 12
As ππ2 β 4ππππ > 0,
So, two distinct real roots exist for this equation as follows:
π₯π₯ =βππ Β± βππ2 β 4ππππ
2ππ
=β(β6) Β± οΏ½(β6)2 β 4(2)(3)
2(2)
=6 Β± β12
4 =6 Β± 2β3
4
=3 Β± β3
2
Hence, the roots are 3+β32
or 3ββ32
.
2. Find the values of ππ for each of the following quadratic equations, so that they have two equal roots.
(i) 2π₯π₯2 + πππ₯π₯ + 3 = 0
(ii) πππ₯π₯(π₯π₯ β 2) + 6 = 0
Solution:
We know that if an equation πππ₯π₯2 + πππ₯π₯ + ππ = 0 has two equal roots, its discriminant (ππ2 β 4ππππ) will be 0.
(i) 2π₯π₯2 + πππ₯π₯ + 3 = 0
Comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we get ππ = 2, ππ = ππ, ππ =3
Discriminant = ππ2 β 4ππππ = (ππ)2 β 4(2)(3) = ππ2 β 24
For equal roots,
Discriminant = 0
ππ2 β 24 = 0
β ππ2 = 24
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β ππ = Β±β24 = Β±2β6
(ii) πππ₯π₯(π₯π₯ β 2) + 6 = 0 or πππ₯π₯2 β 2πππ₯π₯ + 6 = 0
Comparing the equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we get ππ = ππ, ππ =β2ππ, ππ = 6
Discriminant = ππ2 β 4ππππ = (β2ππ)2 β 4(ππ)(6) = 4ππ2 β 24ππ
For equal roots,
π·π· = ππ2 β 4ππππ = 0
4ππ2 β 24ππ = 0
4ππ(ππ β 6) = 0
Either 4ππ = 0 or ππ β 6 = 0
β ππ = 0 or ππ = 6
But, if ππ = 0, then the equation will not have the terms βπ₯π₯2β and βπ₯π₯β.
Hence, if this quadratic equation has two equal roots, then ππ should be 6 only.
3. 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 the breadth of mango grove be π₯π₯.
Length of mango grove will be 2π₯π₯.
Area of mango grove = (2π₯π₯)(π₯π₯) = 2π₯π₯2
Hence, 2π₯π₯2 = 800
β π₯π₯2 =800
2
β π₯π₯2 = 400
Cancelling square on both the sides, we get π₯π₯ = Β±20
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But length cannot be negative.
So, breadth of mango grove = 20 m
Length of mango grove = 2 Γ 20 = 40 m
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 age of one friend be π₯π₯ years.
Age of the other friend will be = (20β π₯π₯) years. (Given in question)
Four years ago, age of 1st friend = (π₯π₯ β 4) years
and, age of 2nd friend = ((20β π₯π₯)β 4) = (16 β π₯π₯) years
As per the question,
(π₯π₯ β 4)(16β π₯π₯) = 48
16π₯π₯ β 64 β π₯π₯2 + 4π₯π₯ = 48
βπ₯π₯2 + 20π₯π₯ β 112 = 0
π₯π₯2 β 20π₯π₯ + 112 = 0
Comparing this equation with πππ₯π₯2 + πππ₯π₯ + ππ = 0, we obtain
ππ = 1,ππ = β20, ππ = 112
Discriminant = ππ2 β 4ππππ = (β20)2 β 4(1)(112)
= 400β 448 = β48
As ππ2 β 4ππππ < 0,
Thus, no real roots are possible for this equation and hence, this situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution:
Let the length and breadth of the rectangular park be ππ and b.
Perimeter = 2(ππ + ππ) = 80
ππ + ππ = 40
or, ππ = 40β ππ
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Area = ππ Γ ππ = ππ(40β ππ) = 40ππ β ππ2
40ππ β ππ2 = 400
ππ2 β 40ππ + 400 = 0
Comparing this equation with ππππ2 + ππππ + ππ = 0, we obtain
ππ = 1,ππ = β40, ππ = 400
Discriminant = ππ2 β 4ππππ = (β40)2 β 4(1)(400)
= 1600β 1600 = 0
As ππ2 β 4ππππ = 0,
Thus, this equation has equal real roots, and hence, this situation is possible.
Root of this equation,
ππ = βππ
2ππ
ππ = β(β40)2(1) =
402 = 20
So, length of park, ππ = 20 m
and breadth of park, ππ = 40 β ππ = 40β 20 = 20 m
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