NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equation. Furthermore, here we’ve provided you with the latest solution for Class 10 CBSE NCERT Maths Chapter 4. Quadratic Equation. As a result here you’ll find solutions to all the exercises. This NCERT Class 10 solution will help you to score good marks in your exam.

What are Quadratic Equations?

In algebra, a quadratic equation is that can be in the form of ax2 + bx + c = 0. Where a is not equals to zero. If a = 0 then the equation will be linear equation. As there will be no ax2 term.

The nature of roots of a quadratic equation ax2 + bx + c = 0 are as follows.
Case 1:
When b2 – 4ac > 0
The equation will have two distinct real roots.

Case 2:
When b2 – 4ac = 0
The equation will have two equal roots

Case 3:
When b2 – 4ac < 0
The equation will have no real roots.

NCERT Solution for Class 10 Maths Chapter 4 – Quadratic Equation Pdf Download

Students can refer to our solution for NCERT Class 10 Maths Chapter 4 – Quadratic Equations. The Chapter 4 Solution of NCERT will help students prepare for the exams and easily crack the exam. Below we’ve provided you with the exercise wise latest solution.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.1

Here you’ll find NCERT Chapter 4 Quadratic Equations Exercise 4.1 Solution.
Exercise 4.1: Solutions of Questions on Page Number: 73

Q1. 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) (iv) (x – 3) (2x + 1) = x(x+5)
(v) (2x – 1)(x-3) = (x+5)(x-1) (vi) x2 + 3x + 1 = (x – 2)2
(vii) (x+2)3 = 2x(x2 – 1) (viii) x3 – 4x – x + 1 = (x – 2)3

Answer: (x + 1)2 = 2(x – 3)
= x2 + 2x + 1 = 2x – 6
= x2 + 7 = 0
It is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(ii) x2 – 2x = (-2)(3 – x)
= x2 – 2x = -6 + 2x
= x2 – 4x + 6 =0
It is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.


(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
x2 – x – 2 = x2 + 2x – 3
3x – 1 = 0
It is not of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.

(iv) (x – 3) (2x + 1) = x(x+5)
= 2x2 – 5x -3 = x2 + 5x
=x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(v) (2x – 1)(x-3) = (x+5)(x-1)
= 2x2 – 7x + 3 = x2 + 4x – 5
x2 – 11x + 8 = 0
It is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

(vi) x2 + 3x + 1 = (x – 2)2
= x2 + 3x + 1 = x2 – 4x + 4
=7x – 3 = 0
It is not of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.

(vii) (x+2)3 = 2x(x2 – 1)
= x3 + 8 + 6x2 + 12x = 2x3 – 2x
= x3 – 14x – 6x2 – 8 = 0
It is not of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation.

(viii) x3 – 4x – x + 1 = (x – 2)3
= x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
= 2x2 – 13x + 9 = 0
It is of the form ax2 + bx + c = 0
Hence, the given equation is a quadratic equation.

Q2. 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 meters) 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)  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.

(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.

Answers:
(i) Let the breadth of the plot be x m. Hence, the length of the plot is (2x + 1) m.
Area of a rectangle = Length x Breadth
∴ 528 = x (2x + 1)
2x2 + x – 528 = 0

(ii) Let the consecutive integers be x and x + 1.
It is given that their product is 306
x(x + 1) = 306
= x2 + x – 306 = 0

(iii) Let Rohan’s age be x. Hence, his mother’s age = x + 26
3 years hence,
Rohan’s age = x + 3
Mother’s age = x + 26 + 3 = x + 29
It is given that the product of their ages after 3 years is 360
(x+3)(x+29) = 360
x2 + 32x +273 = 0

(iv) Let the speed of train be x km/h.
Time taken to travel 480 km = 480/x hrs.
In second condition, let the speed of train = (x – 8) km/hr
It is also given that the train will take 3 hours to cover the same distance.
Therefore, time taken to travel 480 km = (480/x + 3 ) hrs
Speed x Time = Distance

480 + 3x – 3840x – 24 = 480
3x- 3840x = 24
3x2 – 24x – 3840 = 0
x2 – 8x – 1280 = 0

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.2

Here you’ll find NCERT Chapter 4 Quadratic Equations Exercise 4.2 Solution.
Exercise 4.2: Solutions of Questions on Page Number: 76.

Q1. Find the roots of the following quadratic equations by factorization:

Answer:

Roots of this equation are all the values for which (x – 5)(x + 2) = 0
x – 5 = 0 or x + 2 = 0
i.e. x = 5 or x = -2

The roots of this equation are all the values for which (x+2)(2x-3) = 0
x + 2 = 0 or 2x – 3 = 0
i.e x = -2 or x = 3/2.

The roots of this equation are all the values for which = 0

The roots of this equation are the values for which (4x – 1)2 = 0.
Therefore, (4x -1) = 0 or (4x – 1) = 0
i.e x = 1/4 or x = 1/4.

Roots of this equation are the values for which (10x – 1)2 = 0.
Therefore (10x – 1) = 0 or (10x – 1) = 0.
i.e x = 1/10 or x = 1/10

Q2:
(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. 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. Find out the number of toys produced on that day.

Answer:
Let the number of John’s marbles be x.
Therefore, number of Jivanti’s marble = 45 – x
After losing 5 marbles,
Number of John’s marbles = x – 5
Number of Jivanti’s marbles = 45 – x – 5 = 40 – x
It is given that the product of their marbles is 124.

Either x – 36 = 0 or x – 9 = 0
i.e x = 36 or x = 9
If the number of John’s marbles = 36,
Then, number of Jivanti’s marbles = 45 – 36 = 9
If number of John’s marbles = 9,
Then, number of Jivanti’s marbles = 45 – 9 = 36

(ii) Let the number of toys produced be x.
∴ Cost of production of each toy = Rs (55 – x).
It is given that, total production of the toys = Rs 750

Either x – 25 = 0 or x – 30 = 0.
i.e x = 25 or x = 30
Hence, the number of toys will be either 25 or 30.

Q3: Find two numbers whose sum is 27 and the product is 182.
Answer: Let the first number be x and the second number be 27 – x.
Therefore, their product = x (27 – x)
It is given that the product of these numbers is 182.
Therefore, x(27 – x) = 182
x² – 27 x + 182 = 0
x² – 13x – 14x + 182 = 0
x(x – 13) – 14(x – 13) = 0
(x – 13)(x – 14) = 0
Either x – 13 = 0 or x – 14 = 0
i.e., x = 13 or x = 14
If first number = 13, then
Other number = 27 – 13 = 14
If first number = 14, then
Other number = 27 – 14 = 13
Therefore, the numbers are 13 and 14.

Q4: Find two consecutive positive integers, sum of whose squares is 365.
Answer: Let the consecutive positive integers be x and x + 1
Given that x² + (x+1)² = 365
x² + x² + 1 + 2x = 365
2x² + 2x – 364 = 0
x² + x -182 = 0
x² + 14x – 13x – 182=0
x(x + 14) -13(x+14) = 0
(x+14)(x-13) = 0
Either x + 14 = 0 or x – 13 = 0, i.e., x = – 14 or x = 13.
Since the integers are positive, x can only be 13.
∴ x + 1 = 13 + 1 = 14
Therefore, two consecutive positive integers will be 13 and 14.

Q5: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer: Let the base of the right triangle be x cm. Its altitude = (x – 7) cm.
From Pythagoras theorem,
Base² + Altitude = Hypotenuse²
x² + (x – 7)² = 13²
x² + x² + 49 – 14x = 169
2x² – 14x – 120 = 0
x² – 7x – 60 = 0
x² – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0
Either x – 12 = 0 or x + 5 = 0, i.e., x = 12 or x = – 5.
Since sides are positive, x can only be 12.
Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm =5 cm.

Q6: 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 Rs 90, find the number of articles produced and the cost of each article.
Answer: Let the number of articles produced be x.
Therefore, the cost of production of each article = Rs (2x + 3)
It is given that the total production is Rs 90.
x(2x + 3) = 90
2x² + 3x – 90 = 0
2x² + 15x – 12x – 90 = 0
x(2x + 15) -6(2x + 15) = 0
(2x + 15)(x – 6) = 0.
Either 2x + 15 = 0 or x – 6 = 0, i.e., x = -15/2 or x = 6
As the number of articles produced can only be a positive integer, therefore, x can only be 6. Hence, number of articles produced = 6.
Cost of each article = 2 × 6 + 3 = Rs 15.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.3

Here you’ll find NCERT Chapter 4 Quadratic Equations Exercise 4.3 Solution.
Exercise 4.3: Solutions of Questions on Page Number: 87.

Q1: 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+4√3x+3=0 (iv) 2x + x + 4 = 0

Answer:

Q2: Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Answer:

Q1: Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.
(i) 2x2 – 3x + 5 = 0
(ii) 3x² -4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0

Answer:
We know that for a quadratic equation ax2  + bx + c = 0, discriminant is b2 – 4ac.

  1. If b2 – 4ac > 0 → two distinct real roots
  2. If b2 – 4ac = 0 → two equal real roots
  3. If b2 – 4ac < 0 → no real roots (I) 2x2 – 3x + 5 = 0

Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = – 3, c = 5
Discriminant = b2 – 4ac = ( – 3)2 – 4 (2) (5) = 9 – 40
= – 31
As b2  – 4ac < 0,
Therefore, no real root is possible for the given equation.


(ii) 3x² -4√3x + 4 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 3, b = -4√3, c = 4.
Discriminant =b2 – 4ac = (-4√3) – 4(3)(4)
= 48 – 48 = 0
As b2 – 4ac = 0
Therefore, real roots exist for the given equation and they are equal to each other.
And the other roots will be -b/2a and -b/2a

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2x2 – 6x + 3 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 2, b = – 6, c = 3
Discriminant = b2 – 4ac = ( – 6)2  – 4 (2) (3)
= 36 – 24 = 12
As b2  – 4ac > 0,
Therefore, distinct real roots exist for this equation as follows.

Therefore the roots are

Q2: Find the roots of the following equations:

Answer:

Q3: The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.
Answer:
Let the present age of Rehman be x years.
Three years ago, his age was (x – 3) years.
Five years hence, his age will be (x + 5) years.
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.

However, age cannot be negative.
Therefore, Rehman’s present age is 7 years.

Q4: 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.

Answer: Let the marks in Maths be x.
Then, the marks in English will be 30 – x.
According to the given question

If the marks in Maths are 12, then marks in English will be 30 – 12 = 18
If the marks in Maths are 13, then marks in English will be 30 – 13 = 17

Q5: The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Answer: Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m.

However, the side cannot be negative.
Therefore, the length of the shorter side will be 90m.
Hence, length of the larger side will be (90 + 30) m = 120m.

Q6. 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.
Answer: Let the larger and smaller number be x and y respectively.
According to the given question
x² – y² = 180 and y² = 8x
x² – 8x = 180
x² – 8x – 180 = 0
x² – 18x + 10x – 180 = 0
x(x – 18) + 10(x – 18) = 0
(x – 18)(x + 10) = 0
x=18,-10
However, the larger number cannot be negative as 8 times the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only.

Therefore, the numbers are 18 and 12 or 18 and – 12.

Q7: 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.
Answer:

Let the speed of the train be x km/hr.
Time taken to cover 360 km = 360/x hr.
According to the given question,

However, speed cannot be negative
Therefore, the speed of the train is 40km/hr.

Q8: Two water taps together can fill a tank in hours. The tap of a 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.
Answer: Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x – 10) hr
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of tank filled by larger pipe in 1 hour = 1/(x-10)
It is given that the tank can be filled in 9*3/8 = 75/8
hours by both the pipes together.
Therefore,

Time taken by the smaller pipe cannot be  30/8 = 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.

Q9: 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 speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Answer:

Let the average speed of passenger train be x km/h.
Average speed of express train = (x + 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.

Speed cannot be negative. Therefore, the speed of the passenger train will be 33 km/h, and thus, the speed of the express train will be 33 + 11 = 44 km/h.

Q10 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.

Answer: Let the sides of the two squares be x m and y m. Therefore, their perimeter will be 4x and 4y respectively and their areas will be x2 and y2 respectively.
It is given that 4x – 4y = 24
x – y = 6
x = y + 6
Also, x² + y² = 468
(6+y)²+ y²=468
36+ y²+12y+ y² = 468
2y2+12y – 432 = 0
y²+ 6y – 216 = 0
y + 18y – 12y – 216 = 0
y(y + 18) -12(y + 18) = 0
(y+18) (y-12)=0
y=-18 or 12.
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations Exercise 4.4

Here you’ll find NCERT Chapter 4 Quadratic Equations Exercise 4.4 Solution.
Exercise 4.4: Solutions of Questions on Page Number: 91.

Q1: Find the values 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

Answer:
We know that if an equation ax2 + bx + c = 0 has two equal roots, its discriminant (b2 – 4ac) will be 0.
(i) 2x2 + kx + 3 = 0
Comparing equation with ax2 + bx + c = 0, we obtain
a = 2, b = k, c = 3
Discriminant = b2 – 4ac = (k)2 – 4(2) (3)
= k2 – 24
For equal roots, Discriminant = 0
k2 – 24 = 0
k2 = 24

(ii) kx (x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = k, b = – 2k, c = 6
Discriminant = b2 – 4ac = ( – 2k)2  – 4 (k) (6)
= 4k2 – 24k
For equal roots,
b2 – 4ac = 0 4k2 – 24k = 0
4k (k – 6) = 0
Either 4k = 0 or k = 6 = 0
k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only.

Q2: 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.

Answer:
Let the breadth of mango grove be l. Length of mango grove will be 2l.
Area of mango grove = (2l) (l)
= 2l2
2l2 = 800
l2 = 800/ 2 = 400
l2 – 400 = 0
Comparing this equation with al2 + bl + c = 0, we obtain
a = 1 b = 0, c = 400
Discriminant = b2 – 4ac = (0)2 – 4 × (1) × ( – 400) = 1600
Here, b2 – 4ac > 0
Therefore, the equation will have real roots. And hence, the desired rectangular mango grove can be designed
However, length cannot be negative.
Therefore, breadth of mango grove = 20 m
Length of mango grove = 2 × 20 = 40 m

Q3: 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.

Answer:
Let the age of one friend be x years.
Age of the other friend will be (20 – x) years.
4 years ago, age of 1st friend = (x – 4) years
And, age of 2nd friend = (20 – x – 4)
= (16 – x) years
Given that,
(x – 4) (16 – x) = 48
16x – 64 – x2  + 4x = 48
– x2 + 20x – 112 = 0
x2 – 20x + 112 = 0
Comparing this equation with ax2 + bx + c = 0, we obtain
a = 1, b = -20, c = 112
Discriminant = b2 – 4ac = (- 20)2 – 4 (1) (112)
= 400 – 448 = -48
As b2  – 4ac < 0, Therefore, no real root is possible for this equation, and hence, this situation is not possible.

Q4: Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Answer:
Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l+ b = 40 Or, b = 40 – l
Area = l × b = l (40 – l) = 40l – l2
40l – l2 = 400
l2 – 40l + 400 = 0
Comparing this equation with
al2 + bl + c = 0, we obtain
a = 1, b = – 40, c = 400
Discriminate = b2 – 4ac = ( – 40)2 – 4 (1) (400)
= 1600 – 1600 = 0
As b2  – 4ac = 0, Therefore, this equation has equal real roots. And hence, this situation is possible. Root of this equation,

Therefore, length of park, l = 20 m
And breadth of park, b = 40 – l = 40 – 20 = 20 m

NCERT Class 10 Maths All Chapters Solution

Chapter 1: Real Numbers

Chapter 2: Polynomials

Chapter 3: Pair of Linear Equation in Two Variables

Chapter 4: Quadratic Equations

Chapter 5: Arithmetic Progressions

Chapter 6: Triangles

Chapter 7: Coordinate Geometry

Chapter 8: Introduction to Trigonometry

Chapter 9: Some Applications of Trigonometry

Chapter 10: Circles

Chapter 11: Constructions

Chapter 12: Areas Related to Circles

Chapter 13: Surface Areas and Volumes

Chapter 14: Statistics

Chapter 15: Probability