# Chapter 2- Polynomials Exercise 2.2 – NCERT Solutions for class 10 Maths

Chapter 2- Polynomials Exercise 2.2 – NCERT Solutions for class 10 Maths

These solutions are produced by math professionals who have had them evaluated on a regular basis. Students can use these NCERT chapter-by-chapter solutions to study and prepare for their CBSE board exams.

**Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:**

**(i) x**^{2} – 2x – 8

^{2}– 2x – 8

x^{2} − 4x + 2x − 8 = 0

x(x−4) + 2(x−4) = 0

(x−4)(x+2) = 0

(x−4) = 0, (x+2) = 0

x = 4, x = −2 are the zeroes of the polynomial

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** = −coefficient of x / coefficient of x^{2}

**α + β** = −b/a

−2+4 = − (−2)/1

2 = 2

**Product of zeroes** = constant term / coefficient of x^{2}

**α.β **= c/a

−2×4 = −8/1

−8 = −8

Hence, Verified.

**(ii) 4s**^{2} – 4s + 1

^{2}– 4s + 1

4s^{2}−2s−2s+1= 0

2s(2s−1)−(2s−1) = 0

(2s−1)(2s−1) = 0

2s−1= 0, 2s−1 = 0

s =1/2, s = 1/2 are the zeroes of the polynomial.

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** =−coefficient of x / coefficient of x^{2}

**α + β** = −b/a

1/2 + 1/2 = −(−4)/4

1 = 1

**Product of zeroes **= constant term / coefficient of x^{2}

**α.β** = c/a

1/2×1/2 = 1/4

1/4 = 1/4

Hence, Verified.

#### (iii) 6x^{2} – 3 – 7x

6x^{2}−7x−3 = 0

6x^{2}−9x+2x−3 = 0

3x(2x−3)+(2x−3) = 0

(2x−3) = 0, (3x+1) = 0

x = 3/2, x = −1/3 are the zeroes of the polynomial

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** =− coefficient of x / coefficient of x^{2}

**α + β** = −b/a

**α + β** = −(−7)/6

3/2 + −1/3 = (7)/6

7/6 = 7/6

**Product of zeroes **= constant term / coefficient of x^{2}

**α.β** = c/a

3/2 × −1/3 = (−3)/6

(−3)/6 = (−3)/6

(−1)/2 = (−1)/2

Hence, Verified.

#### (iv) 4u^{2} + 8u

4u(u+2) = 0

4u = 0 or u+2 = 0

u = 0 or u = −2 are the zeroes of the polynomial

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** =− coefficient of x / coefficient of x^{2}

**α + β** = −b/a

α + β = −(8)/4

0+(−2) =−2

−2 =−2

**Product of zeroes **= constant term / coefficient of x^{2}

**α.β = c/a**

0 ×−2 = 0/4

0 = 0

Hence, Verified.

#### (v) t^{2} – 15

t^{2}−15 = 0

t = √15

t = −√15, t = +√15 are the zeroes of the polynomial

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** =− coefficient of x / coefficient of x^{2}

**α + β** **=−b/a**

α + β = 0/1

−√15+√15 = 0

0 = 0

**Product of zeroes **= constant term / coefficient of x^{2}

**α.β = c/a**

−√15 ×√15 =−15/1

−15=−15

Hence, Verified.

#### (vi) 3x^{2} – x – 4

3x^{2}−x−4 = 0

3x^{2}−4x+3x−4 = 0

x(3x−4)+(3x−4) = 0

(x+1)(3x−4) = 0

(x+1) = 0 or (3x−4) = 0

x =−1 or x = 4/3 are the zeroes of the polynomial

**Relationship between the zeroes and the coefficients**

**Sum of zeroes** =− coefficient of x / coefficient of x^{2}

α + β =−1/3

−1+4/3=−1/3

1/3=1/3

**Product of zeroes **= constant term / coefficient of x^{2}

α.β = c/a

−1×4/3=−4/3

−4/3=−4/3

Hence, Verified.

#### Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:

**(i) 1/4 , -1**

From the formulas of sum and product of zeroes, we know,

Sum of zeroes = α+β

Product of zeroes = α .β

Sum of zeroes = α+β = 1/4

Product of zeroes = α β = -1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as :-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(1/4)x +(-1) = 0**

**4x ^{2}–x-4 = 0**

**Thus, 4x ^{2}–x–4 is the **quadratic polynomial.

**(ii)**√2, 1/3

Sum of zeroes = α + β =√2

Product of zeroes = α β = 1/3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2} –(**√2

**)x + (1/3) = 0**

**3x ^{2}-3**√2x+1 = 0

**Thus, 3x ^{2}-3**√2x+1

**is the**quadratic polynomial.

**(iii) 0, √5**

Given, Sum of zeroes = α+β = 0

Product of zeroes = α β = √5

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(0)x +**√5

**= 0**

**Thus, x ^{2}+**√5

**is the**quadratic polynomial.

**(iv) 1, 1**

Given, Sum of zeroes = α+β = 1

Product of zeroes = α β = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as :-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–x+1 = 0**

**Thus , x ^{2}–x+1 is the **quadratic polynomial.

**(v) -1/4, 1/4**

Given, Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as :-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x +αβ = 0**

**x ^{2}–(-1/4)x +(1/4) = 0**

**4x ^{2}+x+1 = 0**

**Thus,4x ^{2}+x+1 is the **quadratic polynomial.

**(vi) 4, 1**

Given, Sum of zeroes = α+β =

Product of zeroes = αβ = 1

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

**x ^{2}−(sum of roots)x + product of roots = 0**

**x ^{2}–(α+β)x+αβ = 0**

**x ^{2}–4x+1 = 0**

**Thus, x ^{2}–4x+1 is the **quadratic polynomial.