NCERT Solutions for Class 10 Maths
Chapter 2 Polynomials Exercise 2.2
Page 33
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
(i) x2–2x –8
= x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4–2 = 2 = -(-2)/1
= -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = 4×(-2) = -8 =-(8)/1
= (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1
= 4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -4/4
= -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2)×(1/2) = 1/4
= (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x
= 6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6)
= -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = -(1/3)×(3/2) = -(3/6)
= (Constant term) /(Coefficient of x2 )
(iv) 4u2+8u
= 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4)
= -(Coefficient of u)/(Coefficient of u2)
Product of zeroes = 0×-2 = 0 = 0/4
= (Constant term)/(Coefficient of u2 )
(v) t2–15
= t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)
= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes = √15×(-√15) = -15 = -15/1
= (Constant term) / (Coefficient of t2 )
(vi) 3x2–x–4
= 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)
Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3)
= -(Coefficient of x) / (Coefficient of x2)
Product of zeroes=(4/3)×(-1) = (-4/3)
= (Constant term) /(Coefficient of x2 )
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:-
x2–(α+β)x +αβ = 0
x2–(1/4)x +(-1) = 0
4x2–x-4 = 0
Thus,4x2–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:-
x2–(α+β)x +αβ = 0
x2 –(√2)x + (1/3) = 0
3x2-3√2x+1 = 0
Thus, 3x2-3√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:-
x2–(α+β)x +αβ = 0
x2–(0)x +√5= 0
Thus, x2+√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:-
x2–(α+β)x +αβ = 0
x2–x+1 = 0
Thus , x2–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
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
x2–(-1/4)x +(1/4) = 0
4x2+x+1 = 0
Thus,4x2+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:-
x2–(α+β)x+αβ = 0
x2–4x+1 = 0
Thus, x2–4x+1 is the quadratic polynomial.