Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.1 | Set 2
Last Updated :
23 Jul, 2025
In this section, we delve into Chapter 2 of the Class 10 RD Sharma textbook, focusing on Polynomials. Exercise 2.1 is designed to help students understand the fundamental concepts of polynomials, including their types, properties, and various operations performed on them.
Class 10 RD Sharma Solutions - Chapter 2 Polynomials - Exercise 2.1 | Set 2
This section provides comprehensive solutions for Exercise 2.1 from Chapter 2 of the Class 10 RD Sharma textbook. These solutions aim to assist students in grasping the core principles of polynomials, ensuring they build a solid mathematical foundation for more advanced topics.
Question 11. If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of (α/β) +(β/α)
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = -1/6,
Product of the zeroes =α × β = -1/3.
Now,
(α/β) +(β/α) = (α2 + β2) - 2αβ / αβ
Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
= -25/12
Hence the value of (α/β) +(β/α) is -25/12.
Question 12. If α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of α/β + 2(1/α + 1/β) + 3αβ
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x – 2.
therefore,
Sum of the zeroes = α + β = 6/3
Product of the zeroes = α × β = 4/3
Now,
α/β + 2(1/α + 1/β) + 3αβ = [(α2 + β2) / αβ] + 2(1/α + 1/β) + 3αβ
[ ((α + β)2 - 2αβ) / αβ] + 2(1/α + 1/β) + 3αβ
Now substitute the values of the sum of zeroes and products of the zeroes and we will get,
α/β + 2(1/α + 1/β) + 3αβ = 8
Hence the value of α/β + 2(1/α + 1/β) + 3αβ is 8.
Question 13. If the squared difference of the zeroes of the quadratic polynomial f(x) = x2 + px + 45 is equal to 144, find the value of p.
Solution:
Let as assume that the two zeroes of the polynomial are α and β.
Given that,
f(x) = x2 + px + 45
Now,
Sum of the zeroes = α + β = – p
Product of the zeroes = α × β = 45
therefore,
(α + β)2 - 4αβ = (-p)2 - 4 x 45 = 144
(-p)2 = 144 + 180 = 324
p = √324
Hence the value of p will be either 18 or -18.
Question 14. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – px + q, prove that [(α2 / β2) + (β2 / α2)] = [p4/q2] - [4p2/q] + 2
Solution:
Given that,
α and β are the roots of the quadratic polynomial.
f(x) = x2 – px + q
Now,
Sum of the zeroes = p = α + β
Product of the zeroes = q = α × β
therefore,
LHS = [(α2 / β2) + (β2 / α2)]
= [(α^4 + β4) / α2.β2]
= [((α+ β)^2 - 2αβ)2 + 2(αβ)2] / (αβ)2
= [((p)2 - 2q)2 + 2(q)2] / (q)2
= [(p4 + 4q2 - 4pq2) - 2q2] / q2
= (p4 + 2q2 - 4pq2) / q2 = (p/q)2 - (4p2/q) + 2
LHS = RHS
Hence, proved.
Question 15. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.
Solution:
Given that,
α and β are the zeroes of the quadratic polynomial
f(x) = x2 – p(x + 1)– c
Now,
Sum of the zeroes = α + β = p
Product of the zeroes = α × β = (- p – c)
therefore,
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
= (− p – c) + p + 1
= 1 – c = RHS
therefore, LHS = RHS
Hence proved.
Question 16. If α and β are the zeroes of the quadratic polynomial such that α + β = 24 and α – β = 8, find a quadratic polynomial having α and β as its zeroes.
Solution:
Given that,
α + β = 24 ------(i)
α – β = 8 ------(ii)
By solving the above two equations, we will get
2α = 32
α = 16
put the value of α in any of the equation.
Let we substitute it in (ii) and we will get,
β = 16 – 8
β = 8
Now,
Sum of the zeroes of the new polynomial = α + β = 16 + 8 = 24
Product of the zeroes = αβ = 16 × 8 = 128
Then, The quadratic polynomial = x2– (sum of the zeroes)x + (product of the zeroes) = x2 – 24x + 128
Hence, the required quadratic polynomial is f(x) = x2 + 24x + 128
Question 17. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 1, find a quadratic polynomial whose zeroes are 2α/β and 2β/α.
Solution:
Given that,
f(x) = x2 – 1
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = – 1
therefore,
Sum of the zeroes of the new polynomial
= [(2α2 + 2β2)] / αβ
= [2(α2 + β2)] / αβ
= [2((α + β)2 - 2αβ)] / αβ = 4/(-1)
After substituting the value of the sum and products of the zeroes we will get,
As given in the question,
Product of the zeroes
= (2α)(2β) / αβ = 4
Hence, the quadratic polynomial is
x2 – (sum of the zeroes)x + (product of the zeroes)
= kx2 – (−4)x + 4x2 –(−4)x + 4
Hence, the required quadratic polynomial is f(x) = x2 + 4x + 4
Question 18. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 3x – 2, find a quadratic polynomial whose zeroes are 1/(2α + β) and 1/(2β + α).
Solution:
Given that,
f(x) = x2 – 3x – 2
Sum of the zeroes = α + β = 3
Product of the zeroes = αβ = – 2
therefore,
Sum of the zeroes of the new polynomial
= 1/(2α + β) + 1/(2β + α)
= (2α + β + 2β + α) / (2α + β)(2β + α)
= (3α + 3β) / (2(α2 + β2) + 5αβ)
= (3 x 3) / 2[2(α + β)2 - 2αβ + 5 x (-2)]
= 9 / 2[9-(-4)]-10 = 9/16
Product of zeroes = 1/(2α + β) x 1/(2β + α)
= 1 / (4αβ + 2α2 + 2β2 + αβ)
= 1 / [5αβ + 2((α + β)2 - 2αβ)]
= 1 / [5 x (-2) + 2((3)2 - 2 x (-2))] = 1/16
therefore, the quadratic polynomial is,
x2- (sum of the zeroes)x + (product of the zeroes)
= (x2 + (9/16)x +(1/16))
Hence, the required quadratic polynomial is (x2 + (9/16)x +(1/16)).
Question 19. If α and β are the zeroes of the quadratic polynomial f(x) = x2 + px + q, form a polynomial whose zeroes are (α + β)2 and (α – β)2.
Solution:
Given that,
f(x) = x2 + px + q
Sum of the zeroes = α + β = -p
Product of the zeroes = αβ = q
therefore,
Sum of the zeroes of new polynomial = (α + β)2 + (α – β)2
= (α + β)2 + α2 + β2 – 2αβ
= (α + β)2 + (α + β)2 – 2αβ – 2αβ
= (- p)2 + (- p)2 – 2 × q – 2 × q
= p2 + p2 – 4q = p2 – 4q
Product of the zeroes of new polynomial = (α + β)2 x (α – β)2
= (- p)2((- p)2 - 4q)
= p2 (p2–4q)
therefore, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – (2p2 – 4q)x + p2(p2 – 4q)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2p2 –4q) x + p2(p2 - 4q)).
Question 20. If α and β are the zeroes of the quadratic polynomial f(x) = x2 – 2x + 3, find a polynomial whose roots are:
(i) α + 2, β + 2
(ii) [α-1] / [α+1], [β-1] / [β+1]
Solution:
Given that,
f(x) = x2 – 2x + 3
Sum of the zeroes = α + β = 2
Product of the zeroes = αβ = 3
(i) Sum of the zeroes of new polynomial = (α + 2) + (β + 2)
= α + β + 4 = 2 + 4 = 6
Product of the zeroes of new polynomial = (α + 1)(β + 1)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4 = 3 + 2(2) + 4 = 11
therefore, quadratic polynomial is :
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 – 6x +11
Hence, the required quadratic polynomial is f(x) = k(x2 – 6x + 11).
(ii) Sum of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= [(α-1)(β+1) + (β-1)(α+1)] / (α+1)(β+1)
= [αβ + α - β - 1 + αβ - α + β - 1)] / (α+1)(β+1)
= (3-1+3-1) / (3+1+2) = 2/3
Product of the zeroes of new polynomial :
= [(α-1)/(α+1)] + [(β-1)/(β+1)]
= 26 = 13(2/6) = 1/3
therefore, the quadratic polynomial is,
x2 – (sum of the zeroes)x + (product of the zeroes)
= x2 - (2/3)x + (1/3)
Hence, the required quadratic polynomial is f(x) = k(x2 – (2/3)x + (1/3))
Question 21. If α and β are the zeroes of the quadratic polynomial f(x) = ax2 + bx + c, then evaluate:
(i) α – β
(ii) 1/α - 1/β
(iii) 1/α + 1/β - 2αβ
(iv) α2β + αβ2
(v) α4 + β4
(vi) 1/(aα + b) + 1/(aβ + b)
(vii) β/(aα + b) + α/(aβ + b)
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
Solution:
Given that,
f(x) = ax2 + bx + c
Sum of the zeroes of polynomial = α + β = -b/a
Product of zeroes of polynomial = αβ = c/a
Since, α + β are the zeroes of the given polynomial therefore,
(i) α – β
The two zeroes of the polynomials are :
= [√-b+b2-4ac]/2a - ([-b+√(b2-4ac)]/2a)
= [-b+√(b2-4ac) + b+√(b2-4ac)] / 2a
= √(b2-4ac) / a
(ii) 1/α - 1/β
= (β-1) / αβ = -(α-β)/αβ -------(1)
From above question as we know that,
α-β = √(b2-4ac) / a
and,
αβ = c/a
Put the values in (i) and we will get,
= -[(√(b2-4ac))/c]
(iii) (1/α) + (1/β) - 2αβ
= (α+β)/αβ - 2αβ ---------- (i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= (-b/a x a/c - 2c/a) = -[b/c + 2c/a]
(iv) α2β + αβ2
= αβ(α + β) --------(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/ a
Product of zeroes of polynomial = αβ = c/a
After putting it in (i), we will get
= c/a(-b/a) = -bc/a^2
(v) α4 + β4
= (α2 + β2)2 – 2α2β2
= ((α + β)2 – 2αβ)2 – (2αβ)2 ---------(i)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After substituting it in (i), we will get
= [(-b/a) -2(c/a)]2 - [2(c/a)2]
= [(b2 -2(ac)) / a2]2 - [2(c/a)2]
= [(b2 - 2ac)2 - 2a2 c2] / a4
(vi) 1/(aα + b) + 1/(aβ + b)
= (aβ + b + aα + b) / (aα + b)(aβ + b)
= (a(α + β) + 2b) / (a2 x αβ + abα + abβ + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= b / (ac - b2 + b2) = b/ac
(vii) β/(aα + b) + α/(aβ + b)
= [β(aβ + b) + α(aα + b)] / (aβ + b)(aα + b)
= [aα2 + bβ2 + bα + bβ] / (a2 x (c/a) + ab(α+β) + b2)
Since,
Sum of the zeroes of polynomial = α + β = – b/a
Product of zeroes of polynomial = αβ = c/a
After putting it, we will get
= a[(α+β)2 - b(α+β)] / ac
= a[b2/a - 2c/a] - b2/a
= a[(b2 - 2c - b2)/a] / ac
= (b2 - 2c - b2) / ac = -2/a
(viii) [(α2/β) + (β2/α)] + b[α/a + β/a]
= a[(α2 + β2) / αβ] + b[(α2+β2)/αβ]
= a[(α+β)2 - 2αβ] + b((α+β)2 - 2αβ) / αβ
Since,
Sum of the zeroes of polynomial= α + β = – b/a
Product of zeroes of polynomial= αβ = c/a
After putting it, we will get
= a[(-ba)2 - 3x(c/a)] + b((-b/a)2 - 2(c/a)) / (c/a)
= [(-b2a2/a2c)+(3bca2/a2)+(b/a)2 - (2bca2/a2c)] = b
Summary
Exercise 2.1 | Set 2 of RD Sharma's Class 10 Mathematics textbook focuses on the fundamental concepts of polynomials. This section covers topics such as identifying polynomials, determining their degrees, finding zeros, and evaluating polynomials for given values of variables. Students learn to classify polynomials based on their degrees and number of terms, understand the relationship between factors and zeros, and solve problems involving the behavior of polynomial functions. The exercise also includes questions on finding the values of unknown coefficients in polynomials satisfying certain conditions.
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem