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

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

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
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Knowledge Check

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