show that (α+1)(β+1)=1-c and (α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c)
(α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c)=1
x^2 - p(x+1) - c = 0
x^2 - px - (p+c) = 0
Let the roots be a and b.
a + b = p....(i)
ab = - (p+c).....(ii)
therefore
(1) (a+1)(b+1)
= ab + a + b + 1
= - p - c + p + 1
= 1 - c
also
from (i) and (ii)
c = - (a+b+ab)
now
(2) (a^2+2a+1)/(a^2+2a+c) + (b^2+2b+1)/(b^2+2b+c)
= (a+1)^2/(a^2+2a-a-b-ab) + (b+1)^2/(b^2+2b-a-b-ab)
= (a+1)^2/(a+1)(a-b) + (b+1)^2/(b+1)(b-a)
= (a+1)/(a-b) + (b+1)/(b-a)
= (a+1)/(a-b) - (b+1)/(a-b)
= (a+1-b-1)/(a-b)
= (a-b)/(a-b)
= 1
Hence Solved
You have so many quadratic problems ... : )
x² -p(x + 1) - c = 0 ===> x² -px - (p + c) = 0
α + β = p, αβ = -(p + c),
and (α + 1)(β + 1) = αβ + (α + β) + 1 = 1 - c
(α² + 2α + 1)/(α² + 2α + c)
= (α + 1)²/[(α + 1)² - (1 - c)]
(β² + 2β + 1)/(β² + 2β + c)
= (β + 1)²/[(β + 1)² - (1 - c)]
(α + 1)²/[(α + 1)² - (1 - c)] + (β + 1)²/[(β + 1)² - (1 - c)]
= {(α + 1)²[(β + 1)² - (1 - c)] + (β + 1)²[(α + 1)² - (1 - c)]}/
{[(α + 1)² - (1 - c)][(β + 1)² - (1 - c)]}
= [(1 - c)² - (α + 1)(1 - c) + (1 - c)² - (β + 1)(1 - c)]/
[(1 - c)² - (1 - c)(α + 1 + β + 1) + (1 - c)²]
= [2(1 - c)² - (1 - c)(α + 1 + β + 1)]/[2(1 - c)² - (1 - c)(p + 2)]
= [2(1 - c)² -(1 - c)(p + 2)]/[2(1 - c)² - (1 - c)(p + 2)]
24
if α and β arre roots then
α + β = p and αβ = - c - p ( Viete's relations or Vieta)
then
(α+1)(β+1)= αβ + a + b +1 = p - c -p + 1 = 1-c QED
For
(α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c) I'm waiting for details
x²-p(x+1)-c=0
x²-px-p-c=0
α+β=p/1 =>p
α*β=(-c-p)/1 =>-c-p
(α+1)(β+1) = α*β+α+β+1 =>1-c-p+p =>1-c
(α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c) =???
to what it is supposed to equal to?
JOBERT DO YOU KNOW WHATS BETTER THAN 24
...Twentyfive
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Answers & Comments
Verified answer
x^2 - p(x+1) - c = 0
x^2 - px - (p+c) = 0
Let the roots be a and b.
a + b = p....(i)
ab = - (p+c).....(ii)
therefore
(1) (a+1)(b+1)
= ab + a + b + 1
= - p - c + p + 1
= 1 - c
also
from (i) and (ii)
c = - (a+b+ab)
now
(2) (a^2+2a+1)/(a^2+2a+c) + (b^2+2b+1)/(b^2+2b+c)
= (a+1)^2/(a^2+2a-a-b-ab) + (b+1)^2/(b^2+2b-a-b-ab)
= (a+1)^2/(a+1)(a-b) + (b+1)^2/(b+1)(b-a)
= (a+1)/(a-b) + (b+1)/(b-a)
= (a+1)/(a-b) - (b+1)/(a-b)
= (a+1-b-1)/(a-b)
= (a-b)/(a-b)
= 1
Hence Solved
You have so many quadratic problems ... : )
x² -p(x + 1) - c = 0 ===> x² -px - (p + c) = 0
α + β = p, αβ = -(p + c),
and (α + 1)(β + 1) = αβ + (α + β) + 1 = 1 - c
(α² + 2α + 1)/(α² + 2α + c)
= (α + 1)²/[(α + 1)² - (1 - c)]
(β² + 2β + 1)/(β² + 2β + c)
= (β + 1)²/[(β + 1)² - (1 - c)]
(α + 1)²/[(α + 1)² - (1 - c)] + (β + 1)²/[(β + 1)² - (1 - c)]
= {(α + 1)²[(β + 1)² - (1 - c)] + (β + 1)²[(α + 1)² - (1 - c)]}/
{[(α + 1)² - (1 - c)][(β + 1)² - (1 - c)]}
= [(1 - c)² - (α + 1)(1 - c) + (1 - c)² - (β + 1)(1 - c)]/
[(1 - c)² - (1 - c)(α + 1 + β + 1) + (1 - c)²]
= [2(1 - c)² - (1 - c)(α + 1 + β + 1)]/[2(1 - c)² - (1 - c)(p + 2)]
= [2(1 - c)² -(1 - c)(p + 2)]/[2(1 - c)² - (1 - c)(p + 2)]
= 1
24
if α and β arre roots then
α + β = p and αβ = - c - p ( Viete's relations or Vieta)
then
(α+1)(β+1)= αβ + a + b +1 = p - c -p + 1 = 1-c QED
For
(α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c) I'm waiting for details
x²-p(x+1)-c=0
x²-px-p-c=0
α+β=p/1 =>p
α*β=(-c-p)/1 =>-c-p
(α+1)(β+1) = α*β+α+β+1 =>1-c-p+p =>1-c
(α^2+2α+1)/(α^2+2α+c) + (β^2+2β+1)/(β^2+2β+c) =???
to what it is supposed to equal to?
JOBERT DO YOU KNOW WHATS BETTER THAN 24
...Twentyfive