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2015-04-12T20:16:22+05:30
Given that alpha & beta are the roots of the quadratic equation x^2-2x+3
Comparing it with ax^2+ bx+c, we have a=1, b= -2 & c=3
Therefore, alpha+beta = -(b)/a = -(-2)/1 = 2
& Alpha*Beta = c/a = 3/1 = 3
Now, (alpha+2)+(beta+2) = (alpha+beta)+4 = (2)+4 = 6
& (alpha+2)*(beta+2) = alpha*beta +2(alpha+beta)+4 = 3+2(2)+4 = 3+4+4=11
Hence, the required polynomial is x^2-6x+11
3 4 3
Then if the roots are alpha+beta and beta+2
2015-04-12T22:40:29+05:30

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         f(x) = x² - p x + q
 α and β are the roots of the above equation.
  to find   α² / β² +  β² / α² = ?

   α = [ p + √(p² - 4q)  ]  / 2        and    β = [ p - √(p² - 4 q) ] / 2

so,      α + β = p      and    α β  =  q          and     α² β²  = q²

   =>    α² + β²   =  (α+β)² - 2 αβ  =  p² - 2 q

   =>    α⁴ + β⁴  =  (α² + β²)² - 2 α²β²  =  (p² - 2q)² - 2 q²
                       =  p⁴ - 4 p² q + 4 q² - 2 q²
                       =  p⁴ - 4 p² q + 2 q²
 
NOW ,  α² / β² + β² / α² =  [ α⁴  + β⁴ ] / α² β² = 
                     = [ p⁴ - 4 p² q + 2 q² ] /  q² 

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