If t-1 and -t-1, t belongs to R, are roots of (a+2)x^2+2ax-1=0, then complete set of values of a is

Hi Husen,

Please not that, t-1 and -t-1 add up to -2.

So sum of roots = -2 = (-2a)/(a+2)

or, a/(a+2) = 1............... (which can never be possible for any real value of a)

Hence there can be no value of 'a' for which t-1, -t-1 are roots of that Quadratic Equation.

Hope that was helpful.

All the Best.

Best Regards,

Ashwin (IIT Madras)