**NCERT Solutions for Class 10 Maths**

Chapter 2 Polynomials **Exercise ****2.2**

Chapter 2 Polynomials

**Page 33**

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

**(i) x ^{2} – 2x – 8 (ii) 4s^{2} – 4s + 1 (iii) 6x^{2} – 3 – 7x (iv) 4u^{2} + 8u (v) t^{2} – 15 (vi) 3x^{2} – x – 4**

**(i) x ^{2}–2x –8**

**= **x^{2}– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)

Therefore, zeroes of polynomial equation x^{2}–2x–8 are (4, -2)

Sum of zeroes = 4–2 = 2 = -(-2)/1

= -(Coefficient of x)/(Coefficient of x^{2})

Product of zeroes = 4×(-2) = -8 =-(8)/1

= (Constant term)/(Coefficient of x^{2})

(ii) 4s^{2}–4s+1

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

Therefore, zeroes of polynomial equation 4s^{2}–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -4/4

= -(Coefficient of s)/(Coefficient of s^{2})

Product of zeros = (1/2)×(1/2) = 1/4

= (Constant term)/(Coefficient of s^{2 })

(iii) 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 polynomial equation 6x^{2}–3–7x are (-1/3, 3/2)

Sum of zeroes = -(1/3)+(3/2) = (7/6)

= -(Coefficient of x)/(Coefficient of x^{2})

Product of zeroes = -(1/3)×(3/2) = -(3/6)

= (Constant term) /(Coefficient of x^{2 })

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

= 4u(u+2)

Therefore, zeroes of polynomial equation 4u^{2} + 8u are (0, -2).

Sum of zeroes = 0+(-2) = -2 = -(8/4)

= -(Coefficient of u)/(Coefficient of u^{2})

Product of zeroes = 0×-2 = 0 = 0/4

= (Constant term)/(Coefficient of u^{2 })

(v) t^{2}–15

= t^{2} = 15 or t = ±√15

Therefore, zeroes of polynomial equation t^{2} –15 are (√15, -√15)

Sum of zeroes =√15+(-√15) = 0= -(0/1)

= -(Coefficient of t) / (Coefficient of t^{2})

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

= (Constant term) / (Coefficient of t^{2 })

(vi) 3x^{2}–x–4

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

Therefore, zeroes of polynomial equation3x^{2} – x – 4 are (4/3, -1)

Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3)

= -(Coefficient of x) / (Coefficient of x^{2})

Product of zeroes=(4/3)×(-1) = (-4/3)

= (Constant term) /(Coefficient of x^{2 })

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**

**(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}–(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}–(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}–(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}–(α+β)x +αβ = 0

x^{2}–x+1 = 0

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

(v) -1/4, 1/4

Given,

Sum of zeroes = α+β = -1/4

Product of zeroes = α β = 1/4

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

x^{2}–(α+β)x+αβ = 0

x^{2}–4x+1 = 0

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

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