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Question
If α,β are roots of x2−p(x+1)−c=0, then show that
i. (α+1)(β+1)=1−c
ii. α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1
i. (α+1)(β+1)=1−c
ii. α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=1
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Solution
Given α and β are rate of equation
x2−p(x+1)−c=0
(i) To prove that (α+1)−(β+1)=1−c
Let f(x)=x2−p(x+1)−c=x2−px−p−c=x2−px−(p+c)
Comparing with ax2−bx+c we have
a=1b=−pc=−(p+c)
α+β=−ba=−(p)1=pα.β=ca=−(p+c)1=−(p+c)(α+1).(β+1)=α.β+α+β+1=−(p+c)+p+1=−p−c+p+1=1−c=RHS
Proved
(ii)x2−p(x+1)−c=0 can be written as
X2−px−(p+c)=0αβ are rate
α+β=pαβ=−(p+c)αβ=−p−cc=−(p+α+β)=−α−β−αβα2+2α+1α2+2α−α−1αβ+β2+2β+1β2+2β−α−β−αβα2+α−β−αββ2+β−α−αβα(α+1)−β(α+1)β(β+1)−α(β+1)(α+1)(α−β)(β+1)(β−α)(α+1)(α+1)(α+1)(α−β)+(β+1)(β+1)(β+1)(β−2)=α+1α−β+β+1β−α=α+1α−β+−(β+1)α−β=α+1−β−1α−β=α−βα−β=1=RHS
Given α and β are rate of equation x2−p(x+1)−c=0
(i) To prove that (α+1)−(β+1)=1−c
Let f(x)=x2−p(x+1)−c=x2−px−p−c=x2−px−(p+c)
Comparing with ax2−bx+c we have
a=1b=−pc=−(p+c)
α+β=−ba=−(p)1=pα.β=ca=−(p+c)1=−(p+c)(α+1).(β+1)=α.β+α+β+1=−(p+c)+p+1=−p−c+p+1=1−c=RHS
Proved
(ii)x2−p(x+1)−c=0 can be written as
X2−px−(p+c)=0αβ are rate
α+β=pαβ=−(p+c)αβ=−p−cc=−(p+α+β)=−α−β−αβα2+2α+1α2+2α−α−1αβ+β2+2β+1β2+2β−α−β−αβα2+α−β−αββ2+β−α−αβα(α+1)−β(α+1)β(β+1)−α(β+1)(α+1)(α−β)(β+1)(β−α)(α+1)(α+1)(α+1)(α−β)+(β+1)(β+1)(β+1)(β−2)=α+1α−β+β+1β−α=α+1α−β+−(β+1)α−β=α+1−β−1α−β=α−βα−β=1=RHS
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