find cubic polynomial whose zeros are alpha =2 , bita =1÷3 , gama =1÷2
Answers
Answered by
1
x^3 - (α+β+γ)x^2 + (αβ+βγ+γα)x + αβγ = 0
=> x^3 - (2+(1/3)+(1/2))x^2 + ((2/3)+(1/6)+1)x + 1/3 = 0
=> x^3 - (17/6)x^2 + (11/6)x + (1/3) = 0
Multiply both sides by 6 for simplification
=> 6x^3 - 17x^2 + 11x + 2 = 0
=> x^3 - (2+(1/3)+(1/2))x^2 + ((2/3)+(1/6)+1)x + 1/3 = 0
=> x^3 - (17/6)x^2 + (11/6)x + (1/3) = 0
Multiply both sides by 6 for simplification
=> 6x^3 - 17x^2 + 11x + 2 = 0
nikitasingh79:
chk ur cubic polynomial formula . it is ( - alpha beta gamma) & you put + sign
Answered by
0
Hi there !!
α = 2
β = 1/3
γ = 1/2
Sum of zeros = α + β + γ
= 2 + 1/3 + 1/2
= 2 + 5/6 = 12 + 5/ 6
= 17/6
Sum of the product of zeros taken two at a time = αβ + βγ + αγ
= 2× 1/3 + 1/3 × 1/2 + 2 × 1/2
= 2/3 + 1/6 +1
= 4+1/6 + 1
= 5/6 + 1
= 11/6
Product of zeros = αβγ
= 2 × 1/3 × 1/2
= 1/3
A cubic polynomial is given by :-
k{ x³ - [α + β + γ ]x² + [αβ + βγ + αγ ]x - [αβγ] }
here , k is a constant and can be any no:
k{x³ - 17/6x² + 11/6x - 1/3}
k = 6 ,
6{x³ - 17/6x² + 11/6x - 1/3}
6x³ - 17x² +11x - 2 ------> Required polynomial
α = 2
β = 1/3
γ = 1/2
Sum of zeros = α + β + γ
= 2 + 1/3 + 1/2
= 2 + 5/6 = 12 + 5/ 6
= 17/6
Sum of the product of zeros taken two at a time = αβ + βγ + αγ
= 2× 1/3 + 1/3 × 1/2 + 2 × 1/2
= 2/3 + 1/6 +1
= 4+1/6 + 1
= 5/6 + 1
= 11/6
Product of zeros = αβγ
= 2 × 1/3 × 1/2
= 1/3
A cubic polynomial is given by :-
k{ x³ - [α + β + γ ]x² + [αβ + βγ + αγ ]x - [αβγ] }
here , k is a constant and can be any no:
k{x³ - 17/6x² + 11/6x - 1/3}
k = 6 ,
6{x³ - 17/6x² + 11/6x - 1/3}
6x³ - 17x² +11x - 2 ------> Required polynomial
Similar questions