# Class 10 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.2

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

**(i) x ^{2 }– 2x – 8**

x

^{2 }– 2x – 8 = x^{2 }– 4x + 2x – 8Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the

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free classeswhich will definitely help them in making a wise career choice in the future.= x (x – 4) + 2(x – 4)

= (x – 4) (x + 2)

Therefore, zeroes of equation x

^{2}– 2x – 8 are (4, -2)Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1

i.e. = -(Coefficient of x) / (Coefficient of x

^{2})Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1

i.e.= (Constant term) / (Coefficient of x

^{2})

**(ii) 4s2 – 4s + 1**

4s

^{2 }– 4s + 1 = 4s^{2 }– 2s – 2s +1= 2s(2s – 1) – 1(2s – 1)

= (2s – 1) (2s – 1)

Therefore, zeroes of equation 4s

^{2 }– 4s +1 are (1/2, 1/2)Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4

i.e.= -(Coefficient of s) / (Coefficient of s

^{2})Product of zeros is equal to [(1/2) × (1/2)] = 1/4

i.e.= (Constant term) / (Coefficient of s

^{2})

**(iii) 6x ^{2 }– 3 – 7x**

6x

^{2}– 3 – 7x = 6x^{2}– 7x – 3= 6x

^{2}– 9x + 2x – 3= 3x(2x – 3) + 1(2x – 3)

= (3x + 1) (2x – 3)

Therefore, zeroes of equation 6x

^{2 }– 3 – 7x are (-1/3, 3/2)Sum of zeroes is equal to -(1/3) + (3/2) = (7/6)

i.e.= -(Coefficient of x) / (Coefficient of x

^{2})Product of zeroes is equal to -(1/3) × (3/2) = -(3/6)

i.e.= (Constant term) / (Coefficient of x

^{2})

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

4u

^{2 }+ 8u = 4u(u + 2)Therefore, zeroes of equation 4u

^{2}+ 8u are (0, -2).Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)

i.e. = -(Coefficient of u) / (Coefficient of u

^{2})Product of zeroes is equal to 0 × -2 = 0 = 0/4

i.e. = (Constant term) / (Coefficient of u

^{2})

**(v) t ^{2 }– 15**

t

^{2 }– 15⇒ t

^{2}= 15 or t = ±√15Therefore, zeroes of equation t

^{2}– 15 are (√15, -√15)Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1)

i.e.= -(Coefficient of t) / (Coefficient of t2)

Product of zeroes is equal to √15 × (-√15) = -15 = -15/1

i.e. = (Constant term) / (Coefficient of t2 )

**(vi) 3x ^{2} – x – 4**

3x

^{2}– x – 4 = 3x^{2}– 4x + 3x – 4= x(3x – 4) + 1(3x – 4)

= (3x – 4) (x + 1)

Therefore, zeroes of equation 3x

^{2}– x – 4 are (4/3, -1)Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3)

i.e. = -(Coefficient of x) / (Coefficient of x

^{2})Product of zeroes is equal to (4/3) × (-1) = (-4/3)

i.e. = (Constant term) / (Coefficient of x

^{2})

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

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

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, 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}– (α + β)x + αβ = 0x

^{2 }– (1/4)x +(-1) = 04x

^{2}– x – 4 = 0∴ 4x

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

**(ii) √2, 1/3**

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given Sum of zeroes = α + β =√2

Product of zeroes = αβ = 1/3

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

x

^{2}– (α + β)x + αβ = 0x

^{2}– (√2)x + (1/3) = 03x

^{2 }– 3√2x + 1 = 0∴ 3x

^{2 }– 3√2x + 1 is the quadratic polynomial.

**(iii) 0, √5**

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 0

Product of zeroes = αβ = √5

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

x

^{2}– (α + β)x + αβ = 0x

^{2}– (0)x + √5 = 0∴ x

^{2 }+ √5 is the quadratic polynomial.

**(iv) 1, 1**

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 1

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}– (α + β)x + αβ = 0x

^{2}– x + 1 = 0∴ x

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

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

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = -1/4

Product of zeroes = α β = 1/4

x

^{2}– (α + β)x + αβ = 0x

^{2}– (-1/4)x + (1/4) = 04x

^{2}+ x + 1 = 0∴ 4x

^{2}+ x + 1 is the quadratic polynomial.

**(vi) 4, 1**

Let two zeroes be α, β

∴ Sum of zeroes = α + β

∴ Product of zeroes = αβ

Given, Sum of zeroes = α + β = 4

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}– (α + β)x + αβ = 0x

^{2}– 4x + 1 = 0∴ x

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