Question 1132053
{{{2x^2-2x+5=0 }}}

we need to use  quadratic formula to find roots
and, to determine the nature of the roots we can check if discriminant is  <{{{ 0}}}, 

if {{{b^2 - 4ac < 0 }}} then, as we know,  the roots are{{{ imaginary }}}values

{{{(-2)^2 - 4*2*5 < 0 }}}

{{{4 - 40 < 0 }}}

{{{- 36 < 0 }}}=> the roots are imaginary 


now use  quadratic formula to find roots


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


{{{x = (-(-2) +- sqrt( (-2)^2-4*2*5 ))/(2*2) }}}

{{{x = (2 +- sqrt( 4-36))/4 }}}

{{{x = (2 +- sqrt( -32))/4 }}}

{{{x = (2 +- sqrt( -2*16))/4 }}}

{{{x = (2 +- 4sqrt(2)*i)/4 }}}...simplify


{{{x = (1 +- 2sqrt(2)*i)/2 }}}


roots:

{{{x = (1 +2sqrt(2)*i)/2 }}}

{{{x = (1 - 2sqrt(2)*i)/2 }}}


let {{{alpha= (1 +2sqrt(2)*i)/2 }}} and {{{beta=(1 - 2sqrt(2)*i)/2 }}}

then,

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


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


{{{1/(alpha+1)+(1/beta)=2/(3+2sqrt(2)*i)+(2/(1 - 2sqrt(2)*i ))}}}