Please solve the question in the attachment.... .... ....... .. . ..

the roots cannot be rational (option 3).

the sol is EXTREEEEEMELY HUGE.

first, use GP sum formula to obtain polynomial as (x^k+1)/(x+1). for this to be a polynomial, k needs to be odd. so, p and q are odd.

now consider the eqn 3x^2+px+5q=0. obviously all the coeffs are odd. now we shall prove that ax^2+bx+c=0 can never have rational roots as long as all of a, b and c are odd integers.

assume to the contrary that roots can be rational

now consider the determinant b^2 – 4ac. if roots are rational, this needs to be a square number, say m^2. but m will be odd (since odd – even= odd).

now rewrite it as b^2 – m^2= 4ac

since b is odd and we have just proved that m is odd as well, hence b^2= 8j+1 and m^2= 8i+1 (since any odd number square is always of the form 8k+1, however this would require another proof).

hence b^2 – m^2= 8*t=4ac

or ac= 2t

but since a and c are both odd, their product has to be an odd number, which is a contradiction since it is 2t in the above eqn. hence our initial assumption was wrong and therefore the roots can be anything BUT rational.