if a not equal to b but alpha^2 = 5alptha - 3 and beta^2 = 5 beta - 3 then the equation with roots alpha/beta , beta/alpha is/
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rohitkumargupta
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α² = 5α - 3 and β² = 5β - 3
given,
α² = 5α - 3 and β² = 5β - 3
α² - 5α + 3 = 0 and β² - 5β + 3 = 0
as we know the quadratic formula,
x = [ -b ± √{b² - 4ac} ]/2a
where,
in ( 1 ) a = 1 , b = -5 , c = 3 and in ( 2 ) a = 1 , b = -5 , c = 3
now,
α = [ 5 ±√{25 - 12} ]/2 and β = [ 5 ± √{25 - 12} ] /2
[ given α ≠ β ]
so, let α = [ 5 + √13 ] /2 , β = [ 5 - √13 ]/2
α/β = [5 + √13 ]/2 / [5 - √13 ] /2
(5 + √13) /(5 - √13) * (5 + √13) /(5 + √13)
(5 + √13)²/(25 - 13)
(25 + 13 + 10√3)/12 = (19 + 5√3)/6
→ β/α = [5 - √13]/2 / [5 + √13]/2
(5 - √13)/(5 + √13) * (5 - √13)/(5 - √13)
(5 - √13)²/(25 - 13)
(25 + 13 - 10√13)/(12) = (19 - 5√13)/6
now, sum of zeroes = α/β + β/α
(19 + 5√13)/6 + (19 - 5√13)/6
(19 + 5√13 + 19 - 5√13)/6 = 38/6 = 19/3
product of zeroes = (19 + 5√13)/6 * (19 - 5√13)/6
[19² - (5√13)²]/36 = (361 - 325)/36
36/36 = 1
now, we know the foumula for finding quadratic equations,
x² - (sum of zeroes)x + product of zeroes = 0
x² - (19/3)x + 1 = 0
hence,the equation is,
3x² - 19x + 3 = 0
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