Exercise 2.2 (NCERT) – Polynomials, Class 10 – All Solutions

05 Mar
Class 10, Class 10 - Chapter 2, Class 10 - Polynomials, , , , , ,

Exercise 2.2 (NCERT) - Polynomials, Class 10

Chapter 2 Exercise 2.2 Class 10

Below are the Quick links for all questions of Exercise 2.2, Real Numbers, Class 10.

Click a link to view the solution of corresponding question.

Relationships of Zeros and Coefficients

Splitting the Middle Term

Relationships of Zeros and Coefficients

This exercise is about very basic concepts of Zeros of Polynomials. The zero of a polynomial, P(x), is the value of x for which the value of P(x) is 0. For a quadratic polynomial ax2+bx+c there can be exactly 2 real zeros, say α and β. The relationships between the zeros of the quadratic polynomial and the coefficients of the polynomial are:

𝑃(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐𝑆=Sum of zeros=α+β=−𝑏𝑎𝑃=Product of zeros=α×β=𝑐𝑎

Now if we already know the Sum (S) and Product (P) of a polynomial, the we can directly get the polynomial using the formula written below.

𝑆=Sum of zeros𝑃=Product of zeros𝑡ℎ𝑒𝑛  𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃

Splitting the Middle Term

If you are able to understand the image displayed below then you will solve all questions really easily.

Before we start solving the questions of this exercise, it is very important to understand the concept used to solve these questions – Splitting the Middle Term of a quadratic polynomial. You are suggested to watch the video embedded below completely to understand 🧠 the concept. The tutorial is slightly lengthy so please have some patience in case you want to dive deep into it. 🤿 👀 ❤️ 💃

How it works?

The coefficient of x, b, is split in two parts, say b1 and b2, such that

b1 + b2 = b and

b1 x b2 = a x c

To find the required pair of b1 and b2 you must use prime factorisation of b. 

Do you love Mathematics? If yes, then watch this video to dive deep into the underlying concept.

Below are the solutions of every question of exercise 2.2 of chapter 1, Polynomials, NCERT of class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) \( \begin{aligned} x^2-2x-8 \end{aligned}\)

𝑥2−2𝑥−8=𝑥2−4𝑥+2𝑥−8=𝑥(𝑥−4)+2(𝑥−4)=(𝑥−4)(𝑥+2)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  (𝑥−4)(𝑥+2)=0
⇒  𝑥−4=0  ⇒  𝑥=4  𝑜𝑟  𝑥+2=0𝑜𝑟  𝑥=−2

Answer: 4 & -2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=1,𝑏=−2,𝑐=−8Sum of roots=−𝑏𝑎=−(−2)1=2Sum of roots=α+β=4+(−2)=2𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=−81=−8Product of roots=α×β=4×(−2)=−8𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

(ii) \( \begin{aligned} 4s^2-4s + 1 \end{aligned}\)

4𝑠2−4𝑠+1=4𝑠2−2𝑠−2𝑠+1=2𝑠(2𝑠−1)−1(2𝑠−1)=(2𝑠−1)(2𝑠−1)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  (2𝑠−1)(2𝑠−1)=0
⇒2𝑠−1=0  𝑜𝑟  2𝑠−1=0⇒𝑠=12  𝑜𝑟  𝑠=12

Answer: 1/2 & 1/2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=4,𝑏=−4,𝑐=1Sum of roots=−𝑏𝑎=−(−4)4=1Sum of roots=α+β=12+12=1𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=14Product of roots=α×β=12×12=14𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

(iii) \( \begin{aligned} 6x^2-3-7x \end{aligned}\)

6𝑥2−3−7𝑥=6𝑥2−7𝑥−3=6𝑥2−9𝑥+2𝑥−3=3𝑥(2𝑥−3)+1(2𝑥−3)=(2𝑥−3)(3𝑥+1)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  (2𝑥−3)(3𝑥+1)=0
⇒2𝑥−3=0  ⇒𝑥=32  𝑜𝑟  3𝑥+1=0𝑜𝑟  𝑥=−13

Answer: 3/2 & -1/3 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=6,𝑏=−7,𝑐=−3Sum of roots=−𝑏𝑎=−(−7)6=76Sum of roots=α+β=32–13=9−26=76𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=−36=−12Product of roots=α×β=32×−13=−12𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

(iv) \( \begin{aligned} 4u^2+ 8u \end{aligned}\)

4𝑢2+8𝑢=4𝑢(𝑢+2)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  4𝑢(𝑢+2)=0
⇒4𝑢=0  ⇒𝑢=0  𝑜𝑟  𝑢+2=0𝑜𝑟  𝑢=−2

Answer: 0 & -2 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=4,𝑏=8,𝑐=0Sum of roots=−𝑏𝑎=−84=−2Sum of roots=α+β=0+(−2)=−2𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=04=0Product of roots=α×β=0×(−2)=0𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

(v) \( \begin{aligned} t^2-15 \end{aligned}\)

𝑡2−15=(𝑡+15‾‾‾√)(𝑡−15‾‾‾√)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  (𝑡+15‾‾‾√)(𝑡−15‾‾‾√)=0
⇒𝑡+15‾‾‾√=0  ⇒𝑡=−15‾‾‾√  𝑜𝑟  𝑡−15‾‾‾√=0𝑜𝑟  𝑡=15‾‾‾√

Answer: √15 & -√15 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=1,𝑏=0,𝑐=−15Sum of roots=−𝑏𝑎=01=0Sum of roots=α+β=15‾‾‾√–15‾‾‾√=0𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=−151=−15Product of roots=α×β=15‾‾‾√×(−15‾‾‾√)=−15𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

(vi) \( \begin{aligned} 3x^2-x-4 \end{aligned}\)

3𝑥2−𝑥−4=3𝑥2−4𝑥+3𝑥−4=𝑥(3𝑥−4)+1(3𝑥−4)=(3𝑥−4)(𝑥+1)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔  (3𝑥−4)(𝑥+1)=0
⇒3𝑥−4=0  ⇒𝑥=43  𝑜𝑟  𝑥+1=0𝑜𝑟  𝑥=−1

Answer: 4/3 & -1 are the zeros of the polynomial.

Verification

Comparing the polynomial with ax2 + bx + c

𝑎=3,𝑏=−1,𝑐=−4Sum of roots=−𝑏𝑎=−(−1)3=13Sum of roots=α+β=43−1=13𝐻𝑒𝑛𝑐𝑒  α+β=−𝑏𝑎
Product of roots=𝑐𝑎=−43Product of roots=α×β=43×(−1)=−43𝐻𝑒𝑛𝑐𝑒  α×β=𝑐𝑎

Question 2

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4, -1

𝑆=Sum of zeros=14𝑃=Product of zeros=−1

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−(14)𝑥+(−1)=𝑥2−𝑥4−1

We can stop solving the question further and the polynomial written above is the correct answer. However, if we multiply this polynomial by any non-zero real number, the resulting polynomial will also have the same zeros.

Multiply the P(x) by 4 to remove 4 from the denominator. 

4(𝑥2−𝑥4−1)=4𝑥2−𝑥−4

(ii) √2 , 1/3 

𝑆=Sum of zeros=2‾√𝑃=Product of zeros=13

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−2‾√𝑥+13

Multiply the P(x) by 3 to remove 3 from the denominator. 

3(𝑥2−2‾√𝑥+13)=3𝑥2−32‾√𝑥+1

(iii) 0, √5

𝑆=Sum of zeros=0𝑃=Product of zeros=5‾√

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−0𝑥+5‾√=𝑥2+5‾√

(iv) 1, 1

𝑆=Sum of zeros=1𝑃=Product of zeros=1

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−1𝑥+1=𝑥2−𝑥+1

(v) -1/4, 1/4 

𝑆=Sum of zeros=−14𝑃=Product of zeros=14

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−(−14)𝑥+14=𝑥2+14𝑥+14

Multiply the P(x) by 4 to remove 4 from the denominator. 

4(𝑥2+14𝑥+14)=4𝑥2+𝑥+1

(vi) 4, 1

𝑆=Sum of zeros=4𝑃=Product of zeros=1

Now P(x) = x2 – Sx + P

𝑃(𝑥)=𝑥2−𝑆𝑥+𝑃=𝑥2−4𝑥+1
\( \begin{aligned} \end{aligned}\)