Question 121551
If {{{x-y=9}}} 
and {{{xy=3}}} solve this system to find {{{x}}} and {{{y}}}


{{{ xy=3}}}…..=>.{{{y = 3/x}}}…..=>.substitute into:

{{{ x-y=9}}}…….=>…. 

{{{x-3/x=9}}}...both sides multiply by {{{x}}} 

{{{x^2-3=9x}}} 

{{{x^2 -9x -3=0}}} .....use quadratic formula to solve for {{{x}}}


{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-(-9) +- sqrt ((-9)^2 -4*1*(-3) )) / (2*1)}}}

{{{x[1,2]=(9 +- sqrt (81 + 12 )) / 2}}}

 {{{x[1,2]=(9 +- sqrt (93 )) / 2}}}

{{{x[1,2]=(9 +- 9.64 ) / 2}}}

{{{x[1]=(9 + 9.64 ) / 2}}}

{{{x[1]=(18.64 ) / 2}}}

{{{ highlight (x[1]= 9.32) }}}

{{{x[2]=(9 - 9.64 ) / 2}}}

{{{x[2]=(-.64 ) / 2}}}

{{{ highlight (x[2]= -( .32))}}}


=>. {{{y[1] = 3/x}}} …
{{{ y[1] = (3/9.32)}}}
{{{highlight(y[1]= .32)}}}

=>. {{{y = 3/x}}} …..=>.
{{{  y[2] = (3/-.32)}}}
{{{highlight(y[2] =-9.375) }}}



then 

{{{x^2+y^2= (9.32)^2 + (.32)^2=86.8624 +0.1024=86.9648 = 87}}}... 



{{{x^2+y^2= (- .32)^2 + (-9.375)^2= 0.1024 + 87.890625=87.993025}}}


so, your answer is: 

e) 87