Math, asked by Kanupriya07, 1 year ago

if α is not equals to β but Alpha square is equal to 5α minus 3 beta square equals to 5 α minus 3,beta square is equal to 5 β minus 3 find the equation whose roots are alpha upon beta and beta upon alpha

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Answered by rohitkumargupta
10

HELLO DEAR,





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





I HOPE ITS HELP YOU DEAR,


THANKS

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