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Let alpha!=beta satisfy alpha^(2)+1=6alp...

Let `alpha!=beta` satisfy `alpha^(2)+1=6alpha,beta^(2)+1=6beta.` Then, the quadratic equation whose roots are `(alpha)/(alpha+1),(beta)/(beta+1)` is

A

`8^(2)+8x+1=0`

B

`8x^(2)-8x-1=0`

C

`8x^(2)-8x+1=0`

D

`8x^(2)+8x-1=0`

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