# If alpha and beta are the zeros of th polynomial f(x) =x^2-p(x+1)-c, show that (alpha+beta)(beta+1)=1-c

2
by Sivesh

2015-04-12T20:36:21+05:30
Given that alpha and beta are the roots of the quadratic equation  f(x) = x^2-p(x+1)-c = x^2-px-p-c = x^2 -px-(p+c),
comparing with ax^2 + bx + c, we have, a =1 , b= -p & c= -(p+c)
alpha+beta = -b/a = -(-p)/1 = p
& alpha*beta = c/a = -(p+c)/1 = -(p+c)
Therefore,
(Alpha + 1)*(beta+1)
= Alpha*beta + alpha + beta + 1
= -(p+c) + p + 1
= -p-c+p+1
= 1-c

2015-04-12T22:40:17+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