Question 1041375
{{{2x^2 + x +6 = 0}}} <===> {{{x^2 + x/2 +3 = 0}}} 

Hence {{{alpha+beta = -1/2}}}  and {{{alpha*beta = 3}}}

For the new quadratic equation:


The coefficient of x = -(Sum of the roots)

= -{{{(alpha + 1/(2*beta) + beta + 1/(2*alpha))}}}.

=-{{{(alpha  + beta + 1/(2*beta)+ 1/(2*alpha)) = -(alpha  + beta + 
(1/2)*((alpha+beta)/(alpha*beta)))}}}

= -{{{( -1/2 +(1/2)*((-1/2)/3))}}} = 7/12

The constant in the quadratic equation = product of roots

= {{{(alpha + 1/(2*beta))(beta + 1/(2*alpha)))}}}.

={{{alpha*beta +1 + 1/(4*alpha*beta) = 3+1+1/12 = 49/12}}}.

===> the quadratic equation is {{{x^2+ (7/12)x + 49/12 = 0}}}, or after clearing fractions,

{{{12x^2 + 7x + 49 = 0}}}.