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NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations is a resource created by the team at GFG to help students clear their doubts while solving problems from the NCERT textbook. It helps clear the frustration caused by being stuck on a problem for a while. The NCERT Solutions for Class 10 Maths covers all the questions in the exercise of the NCERT textbook for this chapter.
In this NCERT Class 10 Maths Chapter 4 Solutions, students are introduced to the world of quadratic equations, although quadratic equations are introduced to students in Class 9 under the chapter named "Polynomials" there was no separation of this from polynomials but in reality, there are many applications of quadratic equation, therefore it is studied here in class 10 maths as a stand-alone chapter. In this chapter, students learn about quadratic equations, various methods to solve quadratic equations, the nature of roots based on discriminants for quadratic equations, and many real-life based problems which can be modelled in the form of quadratic equations and then solved to get the required solution.
The Quadratic equations are fundamental in algebra characterized by the highest power of the variable being squared. They are generally written in the form ax2+bx+c=0 where a, b and c are constants and πβ 0. These equations are essential for the solving the various practical problems in the science and engineering. Understanding quadratic equations helps in the analyzing relationships that involve squared terms such as the projectile motion, optimization problems and financial calculations.
These Solutions cover all four exercises of the NCERT Class 10 Maths Chapter 4, which are as follows:
Here, you can also find all of the solutions for the NCERT exercises for CBSE Class 10 Maths.
In this Exercise, problems are based on the basic introductory concept of quadratic equations such as whether a given equation is a quadratic equation or not. Other than this some problems here involve mathematical representation of real-world scenarios which can be modelled in terms of quadratic equations.
(i) (x + 1)2 = 2(x β 3)
Solution:
Here,
LHS = (x + 1)2
= x2 + 2x + 1 (Using identity (a+b)2 = a2 + 2ab + b2)
and, RHS = 2(xβ3)
= 2x β 6
As, LHS = RHS
x2 + 2x + 1 = 2x β 6
x2 + 7 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
(ii) x2 β 2x = (β2) (3 β x)
Solution:
Here,
LHS = x2 β 2x
and, RHS = (β2) (3 β x)
= 2xβ6
As, LHS = RHS
x2 β 2x = 2x β 6
x2 β 4x + 6 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
(iii) (x β 2)(x + 1) = (x β 1)(x + 3)
Solution:
Here,
LHS = (x β 2)(x + 1)
= x2 + (β2+1)x + (β2)(1) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 β x β 2
and, RHS = (x β 1)(x + 3)
= x2 + (β1+3)x + (β1)(3) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 + 2x β 3
As, LHS = RHS
x2 β x β 2 = x2 + 2x β 3
3x β 1 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).
Hence, the equation is equation because highest power of x is 1.
(iv) (x β 3)(2x +1) = x(x + 5)
Solution:
Here,
LHS = (x β 3)(2x +1)
= 2x2 + x +(β3)(2x) + (β3)(1)
= 2x2 β 5x β 3
and, RHS = x(x + 5)
= x2 + 5x
As, LHS = RHS
2x2 β 5x β 3 = x2 + 5x
x2 β 10x β 3 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
(v) (2x β 1)(x β 3) = (x + 5)(x β 1)
Solution:
Here,
LHS = (2x β 1)(x β 3)
= 2x2 + (2x)(β3) +(β1)(x) + (β1)(β3)
= 2x2 β 7x + 3
and, RHS = (x + 5)(x β 1)
= x2 + 5(x) + (β1)(x) + (5)(β1) (Using identity (x+a) (x+b) = x2 + (a+b)x + ab)
= x2 + 4(x) β 5
As, LHS = RHS
2x2 β 7x + 3 = x2 + 4(x) β 5
x2 β 11x + 8 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
(vi) x2 + 3x + 1 = (x β 2)2
Solution:
Here,
LHS = x2 + 3x + 1
and, RHS = (x β 2)2
= x2 β 4x + 4 (Using identity (aβb)2 = a2 β 2ab + b2)
As, LHS = RHS
x2 + 3x + 1 = x2 β 4x + 4
7x β 3 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 because (a=0).
Hence, the equation is equation because highest power of x is 1.
(vii) (x + 2)3 = 2x (x2 β 1)
Solution:
Here,
LHS = (x + 2)3
= x3 + 23 + 3x(2)(x+2) (Using identity (x+y)3 = x3 + y3 + 3xy(x+y))
= x3 + 8 + 6x2 +12x
and, RHS = 2x (x2 β 1)
= 2x3 β 2x
As, LHS = RHS
x3 + 8 + 6x2 +12x = 2x3 β 2x
x3 β 6x2 β 14x β 8 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is not in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 3.
(viii) x3 β 4x2 β x + 1 = (x β 2)3
Solution:
Here,
LHS = x3 β 4x2 β x + 1
and, RHS = (x β 2)3
= x3 β 23 β 3x(2)(xβ2) (Using identity (xβy)3 = x3 β y3 β 3xy(x-y))
= x3 β 8 β 6x2 +12x
As, LHS = RHS
x3 β 4x2 β x + 1 = x3 β 8 β 6x2 +12x
2x2 β 13x + 9 = 0 β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
(i) The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Letβs consider,
Breadth of the rectangular plot = b m
Then, length of the plot = (2b + 1) m.
As, Area of rectangle = length Γ breadth
528 m2 = (2x + 1) Γ x
2x2 + x =528
2x2 + x β 528 = 0 β¦β¦β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x β 528 = 0, which is the required representation of the problem mathematically.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Letβs consider,
The first integer number = x
Next consecutive positive integer will be = x + 1
Product of two consecutive integers = x Γ (x +1)
x2 + x = 306
x2 + x β 306 = 0 β¦β¦β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x β 306 = 0, which is the required representation of the problem mathematically.
(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.
Solution:
Letβs consider,
Age of Rohanβs = x years
So, Rohanβs motherβs age = x + 26
After 3 years,
Age of Rohanβs = x + 3
Age of Rohanβs mother will be = x + 26 + 3 = x + 29
The product of their ages after 3 years will be equal to 360, such that
(x + 3)(x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 β 360 = 0
x2 + 32x β 273 = 0 β¦β¦β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is equation because highest power of x is 2.
Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x β 273 = 0, which is the required representation of the problem mathematically.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Letβs consider,
The speed of train = x km/hr
Time taken to travel 480 km = (480/x) hr
Time = Distance / Speed
Here, According to the given condition,
The speed of train = (x β 8) km/hr
As, the train will take 3 hours more to cover the same distance.
Hence, Time taken to travel 480 km = 480/(x+3) km/hr
As,
Speed Γ Time = Distance
(x β 8)(480/(x + 3) = 480
480 + 3x β 3840/x β 24 = 480
3x β 3840/x = 24
3x2 β 8x β 1280 = 0 β¦β¦β¦β¦β¦β¦.(I)
As, the eqn. (I) is in the form of ax2 + bx + c = 0 where (aβ 0).
Hence, the equation is QUADRATIC equation because highest power of x is 2.
Therefore, the speed of the train, satisfies the quadratic equation, 3x2 β 8x β 1280 = 0, which is the required representation of the problem mathematically.
In this exercise, questions related to the first method which is called splitting the middle term. Other than the basic solution of equations in this exercise some word problems based on real-world scenarios are also included, in which we need to formulate the quadratic equation first and then solve it to find the required solution.
(i) x2β 3x β 10 = 0
Solution:
Here, LHS = x2β 3x β 10
= x2 β 5x + 2x β 10
= x(x β 5) + 2(x β 5)
= (x β 5)(x + 2)
The roots of this equation, x2 β 3x β 10 = 0 are the values of x for which
(x β 5)(x + 2) = 0
Hence, x β 5 = 0 or x + 2 = 0
β x = 5 or x = -2
(ii) 2x2 + x β 6 = 0
Solution:
Here, LHS = 2x2 + x β 6
= 2x2 + 4x β 3x β 6
= 2x(x + 2) β 3(x + 2)
= (2xβ 3)(x + 2)
The roots of this equation, 2x2 + x β 6 = 0 are the values of x for which
(2xβ 3)(x + 2) = 0
Hence, 2xβ 3 = 0 or x + 2 = 0
β x = 3/2 or x = β2
(iii) β2x2 + 7x + 5β2 = 0
Solution:
Here, LHS = β2x2 + 7x + 5β2
= β2x2 + 5x + 2x + 5β2
= x(β2x + 5) + β2(β2x + 5)
= (β2x + 5) (x +β2)
The roots of this equation, β2x2 + 7x + 5β2 = 0 are the values of x for which
(β2x + 5) (x +β2) = 0
Hence, β2x + 5 = 0 or x +β2 = 0
β x = β5/β2 or x = ββ2
(iv) 2x2 β x + 1/8 = 0
Solution:
Here, LHS = 2x2 β x + 1/8
= 1/8(16x2 β 8x + 1)
= 1/8(16x2 β 4x -4x + 1)
= 1/8(4x(4x-1) -1 (4x-1))
= 1/8 (4x-1) (4x-1)
The roots of this equation, 2x2 β x + 1/8 = 0 are the values of x for which
1/8 (4x-1) (4x-1) = 0
(4x-1)2 = 0
Hence, 4x-1 = 0 or 4x-1 = 0
β x = 1/4 or x = 1/4
(v) 100x2 β 20x + 1 = 0
Solution:
Here, LHS = 100x2 β 20x + 1
= 100x2 β 10x β 10x + 1
= 10x(10x β 1) β 1(10x β 1)
= (10x β 1) (10x β 1)
The roots of this equation, 100x2 β 20x + 1 = 0 are the values of x for which
(10x β 1) (10x β 1) = 0
(10x β 1)2 = 0
Hence, 10x β 1 = 0 or 10x β 1 = 0
β x = 1/10 or x = 1/10
(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.
Solution:
Letβs say,
The number of marbles John have = x.
So, number of marble Jivanti have = 45 β x
After losing 5 marbles each,
Number of marbles John have = x β 5
Number of marble Jivanti have = 45 β x β 5 = 40 β x
Here, According to the given condition
(x β 5)(40 β x) = 124
x2 β 45x + 324 = 0
x2 β 36x β 9x + 324 = 0
x(x β 36) -9(x β 36) = 0
(x β 36)(x β 9) = 0
Hence, x β 36 = 0 or x β 9 = 0
x = 36 or x = 9
Therefore,
If, Johnβs marbles = 36, then, Jivantiβs marbles = 45 β 36 = 9
And if Johnβs marbles = 9, then, Jivantiβs marbles = 45 β 9 = 36
(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:
Let us say,
Number of toys produced in a day be x.
Therefore, cost of production of each toy = Rs(55 β x)
Given, total cost of production of the toys = Rs 750
So, x(55 β x) = 750
x2 β 55x + 750 = 0
x2 β 25x β 30x + 750 = 0
x(x β 25) -30(x β 25) = 0
(x β 25)(x β 30) = 0
Hence, x β 25 = 0 or x β 30 = 0
x = 25 or x = 30
Hence, the number of toys produced in a day, will be either 25 or 30.
Solution:
Letβs say,
First number be x and the second number is 27 β x.
Therefore, the product of two numbers will be:
x(27 β x) = 182
x2 β 27x β 182 = 0
x2 β 13x β 14x + 182 = 0
x(x β 13) -14(x β 13) = 0
(x β 13)(x -14) = 0
Hence, x β 13 = 0 or x β 14= 0
x = 13 or x = 14
Hence, if first number = 13, then second number = 27 β 13 = 14
And if first number = 14, then second number = 27 β 14 = 13
Hence, the numbers are 13 and 14.
Solution:
Letβs say,
Two consecutive positive integers be x and x + 1.
Here, According to the given condition,
x2 + (x + 1)2 = 365
x2 + x2 + 1 + 2x = 365
2x2 + 2x β 364 = 0
x2 + x β 182 = 0
x2 + 14x β 13x β 182 = 0
x(x + 14) -13(x + 14) = 0
β (x + 14)(x β 13) = 0
Hence, x β 13 = 0 or x + 14= 0
x = 13 or x = β 14
As, here it is said positive integers, so x can be 13, only.
So,
x = 13
and, x + 1 = 13 + 1 = 14
Hence, two consecutive positive integers will be 13 and 14.
Solution:
Letβs say,
Base of the right triangle be x cm.
So, the altitude of right triangle = (x β 7) cm
Base2 + Altitude2 = Hypotenuse2 (Pythagoras theorem)
x2 + (x β 7)2 = 132
x2 + x2 + 49 β 14x = 169 (using identity (a-b)2 = a2 β 2ab + b2)
2x2 β 14x β 120 = 0
x2 β 7x β 60 = 0 (Dividing by 2)
x2 β 12x + 5x β 60 = 0
x(x β 12) + 5(x β 12) = 0
(x β 12)(x + 5) = 0
Hence, x β 12 = 0 or x + 5= 0
x = 12 or x = β 5
As, here side will be a positive integers, so x can be 12, only.
Therefore, the base of the given triangle is and,
the altitude of this triangle will be (12 β 7) cm =
Solution:
Letβs say,
Number of articles produced be x.
So, cost of production of each article = βΉ (2x + 3)
Here, According to the given condition
x(2x + 3) = 90
2x2 + 3x β 90 = 0
2x2 + 15x -12x β 90 = 0
x(2x + 15) -6(2x + 15) = 0
(2x + 15)(x β 6) = 0
Hence, 2x +15 = 0 or x β 6= 0
x = β15/2 or x = 6
As the number of articles produced can only be a positive integer,
So, x = 6.
Hence, number of articles produced = 6
Cost of each article = 2 Γ 6 + 3 = βΉ 15.
In this exercise, questions are based on the second method for solving quadratic equations which is called completing the square method and also the quadratic formula. Other than this, there are word problems based on real-world scenarios same as the previous exercise.
(i) 2x2β 7x + 3 = 0
Solution:
2x2 β 7x = β 3
Dividing by 2 on both sides, we get
x2 -= -
x2 -2 Γ x Γ= -
On adding ()2 to both sides of equation, we get
(x)2 - 2ΓxΓ+()2 = ()2-
(x-)2 = () β () (Using identity: a2 - 2ab + b2 = (a-b)2)
(x-)2 =
(x-)2 = Β±
x =
x =or x =
x =or x =
x = 3 or x =
(ii) 2x2+ x β 4 = 0
Solution:
2x2 + x = 4
Dividing both sides of the equation by 2, we get
x2 += 2
Now on adding2 to both sides of the equation, we get,
(x)2 + 2 Γ x Γ+ ()2 = 2 + ()2
(x +)2 =(Using identity: a2 + 2ab + b2 = (a+b)2)
x += Β±
x =
x =
Hence, x =or x =
(iii) 4x2 + 4β3x + 3 = 0
Solution:
4x2 + 4β3x = -3
Dividing both sides of the equation by 4, we get
x2 + β3x = -
Now on adding ()2 to both sides of the equation, we get,
(x)2 + 2 Γ x Γ+ ()2 = -+ ()2
(x +)2 = 0 (Using identity: a2 + 2ab + b2 = (a+b)2)
Hence, x = -or x = -
(iv) 2x2+ x + 4 = 0
Solution:
2x2 + x = -4
Dividing both sides of the equation by 2, we get
x2 += -2
Now on adding ()2 to both sides of the equation, we get,
(x)2 + 2 Γ x Γ+ ()2 = - 2 + ()2
(x +)2 =(Using identity: a2 + 2ab + b2 = (a+b)2)
As we know, the square of numbers cannot be negative.
Hence, there is no real root for the given equation, 2x2 + x + 4 = 0.
(i) 2x2β 7x + 3 = 0
Solution:
On comparing the equation with ax2 + bx + c = 0, we get,
a = 2, b = -7 and c = 3
Then roots of the quadratic equation =
x =
x =
x =
x =
x =or x =
x =or
x = 3 or
(ii) 2x2+ x β 4 = 0
Solution:
On comparing the equation with ax2 + bx + c = 0, we get,
a = 2, b = 1 and c = -4
Then roots of the quadratic equation =
x =
x =
x =
x =or x =
(iii) 4x2 + 4β3x + 3 = 0
Solution:
On comparing the equation with ax2 + bx + c = 0, we get,
a = 4, b = 4β3 and c = 3
Then roots of the quadratic equation =
x =
x =
x =
x =or x =
(iv) 2x2+ x + 4 = 0
Solution:
On comparing the equation with ax2 + bx + c = 0, we get,
a = 2, b = 1 and c = 4
Then roots of the quadratic equation =
x =
x =
x =
As we know, the square of a number can never be negative.
Hence, there is no real solution for the given equation.
(i) x2 β 3x -1 = 0, x β 0
Solution:
After rearranging, we get
x2 β 3x -1 = 0
On comparing the equation with ax2 + bx + c = 0, we get,
a = 1, b = -3 and c = -1
Then roots of the quadratic equation =
x =
x =
x =
x =or x =
(ii) , x β -4,7
Solution:
After rearranging,
(x+4)(x-7) = -30
x2 β 3x β 28 = 30
x2 β 3x + 2 = 0
On comparing the equation with ax2 + bx + c = 0, we get,
a = 1, b = -3 and c = 2
Then roots of the quadratic equation =
x =
x =
x =
x =
x =or x =
x = 2 or x = 1
Solution:
Let's take,
Present age of Rehman is x years.
Three years ago, Rehmanβs age was (x β 3) years.
Five years after, his age will be (x + 5) years.
According to the given condition,
3(2x + 2) = (x-3)(x+5)
6x + 6 = x2 + 2x β 15
x2 β 4x β 21 = 0
x2 β 7x + 3x β 21 = 0 (by factorizing)
x(x β 7) + 3(x β 7) = 0
(x β 7)(x + 3) = 0
x = 7, -3
As, age cannot be negative.
Therefore, Rehmanβs present age is 7 years.
Solution:
Let's take,
The marks of Shefali in Maths be x.
Then, the marks in English will be 30 β x.
According to the given condition,
(x + 2)(30 β x β 3) = 210
(x + 2)(27 β x) = 210
-x2 + 25x + 54 = 210
Multiply the equation by (-1),
x2 β 25x + 156 = 0
x2 β 12x β 13x + 156 = 0
x(x β 12) -13(x β 12) = 0
(x β 12)(x β 13) = 0
x = 12, 13
Hence, if the marks in Maths are 12, then marks in English will be 30 β 12 = 18
and, the marks in Maths are 13, then marks in English will be 30 β 13 = 17.
Solution:
Let's take,
Breadth = x
Length = x+30
Diagonal = x+60
Diagonal = β(Length2 + Breadth2)
According to the given condition,
β((x+30)2 + (x)2) = x+60
Squaring both sides,
x2 + (x + 30)2 = (x + 60)2
x2 + x2 + 900 + 60x = x2 + 3600 + 120x
x2 β 60x β 2700 = 0
x2 β 90x + 30x β 2700 = 0
x(x β 90) + 30(x -90) = 0
(x β 90)(x + 30) = 0
x = 90, -30
As, side of the field cannot be negative.
Hence, the length of the shorter side will be 90 m.
and, the length of the larger side will be (90 + 30) m = 120 m.
Solution:
Let's take,
The larger number = x
and, smaller number = y
According to the given condition,
x2 β y2 = 180 and y2 = 8x (It means x has to be positive, because it is obtained by squaring a number)
x2 β 8x = 180
x2 β 8x β 180 = 0
x2 β 18x + 10x β 180 = 0
x(x β 18) +10(x β 18) = 0
(x β 18)(x + 10) = 0
x = 18, -10
As x cannot be negative,
Hence, the larger number will be 18.
x = 18
So, As y2 = 8x
= 8 Γ 18
= 144
y = Β±β144 = Β±12
So, Smaller number = Β±12
Hence, the numbers are 18 and 12 or 18 and -12.
Solution:
Let's take
The speed of the train = x km/hr.
As, Speed =
Time taken to cover 360 km =hr.
As per the question given,
(x + 5)(- 1) = 360
(x + 5)() = 360
(x + 5)(360 - x) = 360x
x2 + 5x -1800 = 0
x2 + 45x - 40x + 1800 = 0
x(x + 45) -40(x + 45) = 0
(x + 45)(x β 40) = 0
x = 40, -45
As we know, the value of speed cannot be negative.
Hence, the speed of train is 40 km/h.
Solution:
Let's take
The time taken by the smaller pipe to fill the tank = x hr.
Time taken by the larger pipe = (x β 10) hr
Part of tank filled by smaller pipe in 1 hour =
Part of tank filled by larger pipe in 1 hour =
According to the given condition,
9hrs taken to fill with both the pipe.
So,
75(2x β 10) = 8x2 β 80x
150x β 750 = 8x2 β 80x
8x2 β 230x +750 = 0
8x2 β 200x β 30x + 750 = 0
8x(x β 25) -30(x β 25) = 0
(x β 25)(8x -30) = 0
x = 25,
Time taken by the smaller pipe cannot behours, as the time taken by the larger pipe will become negative.
Hence, time taken by the smaller pipe = 25hours
and, by the larger pipe =15 hours
Solution:
Let's take
The average speed of passenger train = x km/h.
Average speed of express train = (x + 11) km/h
According to the given condition,
= 1
= 1
= 1
132 Γ 11 = x(x + 11)
x2 + 11x β 1452 = 0
x2 + 44x -33x -1452 = 0
x(x + 44) -33(x + 44) = 0
(x + 44)(x β 33) = 0
x = β 44, 33
As we know, Speed cannot be negative.
Hence, the speed of the passenger train will be 33 km/h
and, the speed of the express train will be 33 + 11 = 44 km/h.
Solution:
Let the sides of the two squares be x and y meter.
Perimeter = 4x and 4y respectively
Area = x2 and y2 respectively.
According to the given condition,
4x β 4y = 24
x β y = 6
x = y + 6 ......................(I)
and,
x2 + y2 = 468
(6 + y)2 + y2 = 468 (From (I))
36 + y2 + 12y + y2 = 468
2y2 + 12y + 432 = 0
y2 + 6y β 216 = 0
y2 + 18y β 12y β 216 = 0
y(y +18) -12(y + 18) = 0
(y + 18)(y β 12) = 0
y = -18, 12
As we know, the side of a square cannot be negative.
Hence, the sides of the squares are 12 m
and, (12 + 6) m = 18 m.
This exercise contains problems based on the nature of the root of the quadratic equation which can be determined using the quadratic formula. Other than that same word problems are based on real-world scenarios.
(i) 2x2-3x+5=0
Solution:
(i) Given: 2x2-3x+5=0
Here a=2,b=-3 and c=5
Discriminant, D=b2-4ac
β D = (-3)2β 4 Γ 2 Γ 5)
β D = 9-40 = -31 < 0
Hence, the roots are imaginary.
(ii) 3x2-4β3x+4=0
Solution:
(ii) Given: 3x2-4β3x + 4 = 0
Here a=3,b=β3 and c=4
Discriminant, D=b2-4ac
β D = (-4β3)2 β (4 Γ 3 Γ 4)
β D= 48 β 48 = 0
Hence, the roots are real and equal.
Using the formula,
, we get
Hence, the equal roots are and .
(iii) 2x2-6x+3=0
Solution:
(iii) Given: 2x2-6x+3=0
Here, a=2,b=-6 and c=3
Discriminant, D=b2-4ac
β D = (-6)2 β (4 Γ 2 Γ 3)
β D = 36 β 24 = 12 > 0
Hence, the roots are distinct and real.
Using the formula,
,we get
Hence, the equal roots are and
(i)2x2+kx+3
Solution:
This equation is of the form ax2+bx+x, where a=2, b=k and c=3.
Discriminant, D=b2-4ac
βD= k2 β 4 Γ 2 Γ 3
βD = k2 -24
For equal roots D=0
β k2-24=0
β k2=24
β k2 = Β±24 = Β±2β6β
(ii) kx(x-2)+6=0
Solution:
β kx2-2kx+6=0
This equation is of the form ax2+bx+c=0, where a=k, b=-2k and c=6.
Discriminant, D=b2-4ac
β D =(-2k)2 β 4 Γ k Γ 6
β D =4k2-24k
For equal roots D=0
β 4k2-24k=0
β 4k(k-24)=0
β k=0 (not possible) or 4k-24=0
β k= 24/4=6
Solution:
Let the breadth of the rectangular mango grove be x m.
Then, the length of the rectangular mango grove will be 2x m.
The Area of the rectangular mango grove=length Γ breadth
According to the question, we have
x Γ 2x= 800
β 2x2=800
β x2=400
β x=20
Hence, the rectangular mango grove is possible to design whose length=40 m and breadth=20 m.
Solution:
Let the present age of one friend be x years.
Then, the present age of other friend be (20-x) years.
4 years ago, one friendβs age was (x-4) years
4 years ago, other friendβs age was (20-x-4)=(16-x) years.
According to the question,
(x-4)(16-x)=48
β 16x-64-x2+4x=48
β x2-20x+112=0
This equation is of the form ax2+bx+c=0,where a=1, b=-20 and c=112.
Discriminant, D=b2-4ac
β D = (-20)2-4 Γ 1 Γ 112 = -48 < 0
Since, there are no real roots.
So the given situation is not possible.
Solution:
Let the length of the rectangular park be x.
The perimeter of the rectangular park= 2(length + breadth)
β 2(x + breadth)=80
β breadth=40-x
The area of rectangular park= length Γ breadth
β x(40-x)=400
β 40x-x2=400
β x2-40x+400=0
β x2 -20x-20x+400=0
β (x-20)(x-20)=0
β x=20
Hence, the rectangular park is possible to design. So, the length of the park is 20m and the breadth = 40-20=20m.
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