Verified answer

Given equation:

x^2-p(x+1)-c=0

=>x^2 - px - p - c = 0

=>x^2 - px - (p + c) = 0

Sum of roots = α + β = -b/a = p/1 = p

Product of roots = αβ = c/a = -(p + c)/1 = - p - c

=>p = - αβ - c

=>α + β = -αβ - c

=>c = -α - β - αβ..........i)

To prove: (α+1)(β+1)=1-c

L.H.S. = (α+1)(β+1) = αβ + α + β + 1 = αβ + (α + β) + 1

=p - p - c + 1 = 1 - c = R.H.S.(Proved)

To prove: (α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c)=1

L.H.S. = (α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c)

=(α+1)^2/(α^2+2α -α - β - αβ) + (β+1)^2/(β^2+2β -α - β - αβ)........[From i)]

=(α+1)^2/(α^2 + α - β - αβ) + (β+1)^2/(β^2+β -α - αβ)

=(α+1)^2/(α^2 + α - αβ - β) + (β+1)^2/(β^2+β - αβ - α)

=(α+1)^2/{α(α+1) -β(α+1)} + (β+1)^2/(β(β+1) -α(β + 1)}

=(α+1)^2/{(α+1)(α-β)} + (β+1)^2/((β+1)(β-α)}

=(α+1)/(α-β) + (β+1)/(β-α)

=(α+1)/(α-β) - (β+1)/(α-β)

=(α+1-β-1)/(α-β)

=(α-β)/(α-β)

=1=R.H.S.(Proved)