Question

if alpha is not equal to beta but alpha square=2 alpha-3, beta square =2 beta then the equation whose roots are alpha÷beta and beta÷alpha
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Answer 1

Answer:

Step-by-step explanation:

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