NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 2 Polynomials Class 10 NCERT Solutions Ex 2.4. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 2
Chapter NamePolynomials
ExerciseEx 2.4
Number of Questions Solved5
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4

Question 1.Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 27

Solution:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 28
On comparing the given polynomial with the polynomial ax3 + bx2 + cx + d, we obtain a = 2, b = 1, c = -5, d = 2
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 29
Thus, the relationship between the zeroes and the coefficients is verified.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 30
On comparing the given polynomial with the polynomial ax3 + bx2 + cx + d, we obtain a = 1, b = -4, c = 5, d = -2.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 31
Thus, the relationship between the zeroes and the coefficients is verified.

Concept Insight: The zero of a polynomial is that value of the variable which makes the polynomial 0. Remember that there are three relationships between the zeroes of a cubic polynomial and its coefficients which involve the sum of zeroes, product of all zeroes and the product of zeroes taken two at a time.

Question 2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

Solution:
Let the polynomial be ax3  + bx2 + cx + d and its zeroes be α, β, γ.
According to the given information:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 32
If a = k, then b = -2k, c = -7k, d = 14k
Thus, the required cubic polynomial is k(x3 – 2x2 – 7x + 14), where k is any real number.
The simplest polynomial will be obtained by taking k = 1.

Concept Insight: A cubic polynomial involves four unknowns and we have three relations involving these unknowns, so the coefficients of the cubic polynomial can be found by assuming the value of one unknown and then finding the other three unknowns from the three relations.

Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are  a – b, a, a + b, find a and b.

Solution:
Let p(x) = x3 – 3x2 + x + 1
The zeroes of the polynomial p(x) are given as a – b, a, a + b.
Comparing the given polynomial with dx3 + ex2 + fx + g, we can observe that
d = 1, e = -3, f = 1, g = 1
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 33
Therefore, the zeroes of p(x) are 1-b, 1, 1+b .
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 34

Concept Insight: When the polynomial and its zeroes are given then, remember to apply relationships between the zeroes and coefficients. These relations involve the sum of the zeroes, product of zeroes and the product of zeroes taken two at a time.

Question 4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ±√3, find other zeroes.

Solution:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 35
Therefore, x2 – 4x + 1 is a factor of the given polynomial. For finding the remaining zeroes of the given polynomial, we will carry out the division of
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 36
Hence, 7 and -5 are also zeroes of the given polynomial.

Concept Insight: If a is a zero of a polynomial p(x), where degree of p(x) is greater than 1, then (x – a) will be a factor of p(x) that is when p(x) is divided by (x – a), then the remainder obtained will be 0 and the quotient will be a factor of p(x) using division algorithm.  Remember that if (x – a) and (x – b) are factors of a polynomial, then (x – a)(x – b) will also be a factor of that polynomial.

Question 5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:
By division algorithm,
Dividend = Divisor x Quotient + Remainder
Divisor x Quotient = Dividend – Remainder
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 37
It will be perfectly divisible by x2 – 2x + k.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 38
10 – a – 8 x 5 + 25 = 0
10 – a – 40 + 25 = 0
– 5 – a = 0
a = -5
Thus, k = 5 and a = -5.

Concept Insight:  When a polynomial is divided by any non-zero polynomial, then it satisfies the division algorithm which is as below:
Dividend = Divisor x Quotient + Remainder
Divisor x Quotient = Dividend – Remainder
So, from this relation, it can be said that the result (Dividend – Remainder) will be completely divisible by the divisor.

We hope the NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 2 Polynomials Class 10 NCERT Solutions Ex 2.2. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 2
Chapter NamePolynomials
ExerciseEx 2.2
Number of Questions Solved2
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2

Page No: 33

Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 x – 4

Solution:

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 2
So, the zeroes of x² – 2x – 8 are 4 and -2.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 3
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 4
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 5
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 6
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 7
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 8

Concept Insight: The zero of a polynomial is that value of the variable which when substituted in the polynomial makes its value 0. When a quadratic polynomial is equated to 0, then the values of the variable obtained are the zeroes of that polynomial. The relationship between the zeroes of a quadratic polynomial with its coefficients is very important. Also, while verifying the above relationships, be careful about the signs of the coefficients.

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1
(ii) √2, 1/3
(iii) 0, √5
(iv) 1,1
(v) -1/4,1/4
(vi) 4,1

Solution:
(i)    Let the required polynomial be  ax² + bx + c, and let its zeroes α and β

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 9
If a = 4k, then b = -k, c = -4k  Therefore, the quadratic polynomial is k(4 x 2 – x – 4), where k is a real number .
(ii)     Let the polynomial be  ax² + bx + c, and let its zeroes be α and β
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 10
(iii)    Let the polynomial be  ax² + bx + c, and let its zeroes be α and β
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 11
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 12
(iv)    Let the polynomial be  ax² + bx + c, and let its zeroes be α and β
Therefore, the quadratic polynomial is k(x² – x + 1), where k is a real number.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 13
(v)    Let the polynomial be ax² + bx + c, and its zeroes be α and β
Therefore, the quadratic polynomial is k(4x² + x + 1),where k is a real number .
(vi)    Let the polynomial be ax² + bx + c.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials 14
Therefore, the quadratic polynomial is k(x² – 4x + 1), where k is a real number.

Concept Insight:  Since the sum and product of zeroes give 2 relations between three unknowns so we assign a value to the variable a and obtain other values.
Alternatively, If the sum and the product of the zeroes of a quadratic polynomial is given then polynomial is given by, where k is a constant. And the simplest polynomial will be the one in which k = 1.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 1 Real Numbers Class 10 NCERT Solutions Ex 1.3. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 1
Chapter NameReal Numbers
ExerciseEx 1.3
Number of Questions Solved3
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Page No: 14

Question 1
Prove that √5 is irrational.

Solution:
Let us assume, on the contrary, that √5 is a rational number.
Therefore, we can find two integers a,b (b # 0) such that √5 = a\b
Where a and b are co-prime integers.
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 11
This means that b2 is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
So our assumption that √5 is rational wrong.
Hence,√5 cannot be a rational number. Therefore, √5 is rational.

You can also Download Class 10 Maths NCERT Solutions to help you to revise complete Syllabus and score more marks in your examinations.

Concept Insight: There are various ways of proving in mathematics proof by contradiction is one of them. In this approach, we assume something which is contrary to what needs to be proved and arrive at a fact which contradicts something which is true in general. Key result used here is “If P is a prime number and it divides a2then it divides a as well”.

Question 2
Prove that 3 + 2√5 is irrational.

Solution:
Let us assume, on the contrary that 3 + 2√5 is rational,
Therefore, we can find two integers a, b (b ≠ 0) such that
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 13

Concept Insight: This problem is solved using proof by contradiction. The key concept used is if p is a prime number then √P is irrational. Do not prove this question by assuming sum of rational and irrational is irrational.

Question 3
Prove that the following are irrationals:
(i) 1/√2
(ii) 7√5
(iii) 6 + √2

Solution:

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 14
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 15
NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 16

Concept Insight: This problem is solved using proof by contradiction. The key concept used is if p is a prime number then √P is irrational. Do not prove this question by assuming sum or product of rational and irrational is irrational.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 1 Real Numbers Class 10 NCERT Solutions Ex 1.2. 

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 1
Chapter NameReal Numbers
ExerciseEx 1.2
Number of Questions Solved7
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Page No: 11

Question 1
Express each number as product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Solution:
maths class 10 ncert solutions chapter 1 Real numbers

Concept Insight:   Since the given number needs to be expressed as the product of prime factors so in order to solve this problem knowing prime numbers is required. Do not forget to put the exponent in case a prime number is repeating.

You can also Download Class 10 Maths NCERT Solutions to help you to revise complete Syllabus and score more marks in your examinations.

Question 2
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Solution:
E:\image\NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 6.png

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers 7

Concept Insight: HCF is the product of common prime factors raised to least power, while LCM is product of prime factors raised to the highest power. HCF is always a factor of the LCM.
Do not skip verification product of two numbers = HCF x LCM as it can help in cross checking the answer.

Question 3
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25

Solution:

Class 10 Maths Chapter 1 Real Numbers 9

Concept Insight: HCF is the product of common prime factors of all three numbers raised to least power, while LCM is a product of prime factors of all here raised to the highest power.  Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c)  can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .

Question 4
Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

Maths Chapter 1 Real Numbers Class 10

Concept Insight: This problem must be solved using a product of two numbers = HCF x LCM rather then prime factorization

Question 5
Check whether 6n can end with the digit 0 for any natural number n.

Solution:
If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorization must include primes 2 and  5 both Prime factorization of 6n = (2 x 3)n. By Fundamental Theorem of Arithmetic Prime factorization of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.

Concept Insight: In order to solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorization of a number is unique.

Question 6
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:
Numbers are of two types – prime and composite. Prime numbers have only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself.
It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6
The given expression has 6 and 13  as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x  4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.

Question 7
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:
It can be observed that Ravi and Sonia do not take the same amount of time Ravi takes lesser time than Sonia for completing 1 round of the circular path.
As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
i.e When  Sonia completes one round then ravi completes 1.5 rounds.   So they will meet first time at the time which is a common multiple of the time taken by them to complete 1 round
i.e LCM of 18 minutes and 12 minutes.
Now 18 = 2 x 3 x 3 = 2 x 32
And, 12 = 2 x 2 x 3 = 22 x 3
LCM of 12 and 18 = product of factors raised to highest exponent = 22 x 32 = 36
Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Concept Insight: In order to solve the word problems first step is to interpret the problem and identify what is to be determined. The problem asks for simultaneous reoccurrence of events so we need to find LCM. The key word for simultaneous reoccurrence of events is LCM. Do not forget to write the final answer.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2, drop a comment below and we will get back to you at the earliest.