find the quadratic equation whose roots are 7 + root 3 and 7 minus root 3
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Answered by
34
sum of zeroes = 7+√3+7-√3= 14
product of zeroes = (7+√3)(7-√3)= (7)^2 - (√3)^2
product of zeroes= 49-3= 46
quadratic eq= x^2- ( sum of zeroes)x + ( product of zeroes)
= x^2 - 14x+46
product of zeroes = (7+√3)(7-√3)= (7)^2 - (√3)^2
product of zeroes= 49-3= 46
quadratic eq= x^2- ( sum of zeroes)x + ( product of zeroes)
= x^2 - 14x+46
Answered by
58
Hi there !!
α = 7 +√3
β = 7 - √3
Sum of zeros = α + β = 7 +√3 + 7 -√3
= 14
Product of zeros = αβ = (7 +√3)(7 -√3)
= 7² - 3 = 46
Quadratic equation :-
x² - (α + β)x + αβ
= x² - 14x + 46 ----> required quadratic equation
α = 7 +√3
β = 7 - √3
Sum of zeros = α + β = 7 +√3 + 7 -√3
= 14
Product of zeros = αβ = (7 +√3)(7 -√3)
= 7² - 3 = 46
Quadratic equation :-
x² - (α + β)x + αβ
= x² - 14x + 46 ----> required quadratic equation
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