Question 655375
{{{x1}}} and {{{x2}}} are two solutions of the equation {{{x^2+bx+c=0}}}


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}...since {{{a=1}}}, we have


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


so, the solutions are:


{{{x_1 = (-b +sqrt( b^2-4c ))/2 }}}


{{{x_2 = (-b - sqrt( b^2-4c ))/2 }}}




now find {{{x_1+x_2}}}  and  {{{x_1 *_2}}}


{{{x_1+x_2=(-b +sqrt( b^2-4c ))/2+(-b - sqrt( b^2-4c ))/2}}} 


{{{x_1+x_2=(-b +sqrt( b^2-4c )-b+sqrt( b^2-4c ))/2}}} 


{{{x_1+x_2=(-2b +2sqrt( b^2-4c ))/2}}} 


{{{x_1+x_2=2(-b +sqrt( b^2-4c ))/2}}} 


{{{x_1+x_2=cross(2)(-b +sqrt( b^2-4c ))/cross(2)}}} 


{{{x_1+x_2=-b +sqrt( b^2-4c )}}} 



and

{{{x_1 *_2=((-b +sqrt( b^2-4c ))/2)*((-b - sqrt( b^2-4c ))/2)}}}



 {{{x_1 *_2=(-b +sqrt( b^2-4c ))*(-b - sqrt( b^2-4c ))/(2*2)}}}


{{{x_1 *_2=(b^2 +b*sqrt( b^2-4c )-b*sqrt( b^2-4c ) +( sqrt( b^2-4c ))^2)/4}}}


{{{x_1 *_2=(b^2 +cross(b*sqrt( b^2-4c ))-cross(b*sqrt( b^2-4c )) +( b^2-4*c ))/4}}}


{{{x_1 *_2=(b^2 + b^2-4c )/4}}}


{{{x_1 *_2=(2b^2-4c )/4}}}


{{{x_1 *_2=2(b^2-2c )/4}}}


{{{x_1 *_2=cross(2)(b^2-2c )/cross(4)}}}


{{{x_1 *_2=(b^2-2c )/2}}}