CLASS-10 MATHS POLYNOMIALS EX-2.2 Solutions….
Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4
Solution:
(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 x2 )
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of x2 )
(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)/(Coefficientof 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)/(Coefiicient of u2 )
(v) t 2–15
⇒ 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 equation 3x2 – 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)
Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of zeroes respectively:
Solution:
(i) 1/4 , -1
Solution:
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 – (α + β ) x +αβ = 0
x – (1/4) x + (-1) = 0
4x –x – 4 = 0
Thus, 4x – x–4 is the quadratic polynomial.
(ii) √2, 1/3
Solution:
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
Solution:
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
Solution:
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 + 1 is the quadratic polynomial.
(v) -1/4, 1/4
Solution:
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.