Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2–2x –8
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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 )
Q2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
Sol:
From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α+β
Product of zeroes = αβ
Sum of zeroes = α+β = 1/4
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:
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
Sol:
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:
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
Sol:
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
Sol:
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
Sol:
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
Sol:
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:
x2–(α+β)x+αβ = 0
x2–4x+1 = 0
Thus, x2–4x+1 is the quadratic polynomial.
Check out the NCERT Solutions of all the exercises of Polynomials:
Exercise 2.1. NCERT Solutions: Polynomials
1. What is a polynomial? |
2. What are the different types of polynomials? |
3. How can I identify the degree of a polynomial? |
4. What is the zero of a polynomial? |
5. How can I divide polynomials using long division? |
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