Q.1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Solutions:
(i) x^{2}–2x –8
⇒ x^{2}– 4x+2x–8 = x(x–4)+2(x–4) = (x4)(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(2s1) = (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)(2x3)
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(3x4)+1(3x4) = (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 })
Q.2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(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^{2}–(α+β)x +αβ = 0
x^{2}–(1/4)x +(1) = 0
4x^{2}–x4 = 0
Thus,4x^{2}–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:
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
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:
x^{2}–(α+β)x +αβ = 0
x^{2}–(0)x +√5= 0
Thus, x^{2}+√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:
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
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:
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
Solution:
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}–(α+β)x+αβ = 0
x^{2}–4x+1 = 0
Thus, x^{2}–4x+1 is the quadratic polynomial.
Check out the NCERT Solutions of all the exercises of Polynomials:
Exercise 2.1. NCERT Solutions: Polynomials
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