Question 177583
*[tex \huge x + y = 12 \text { }\Rightarrow\text { }\math x=12 - y]


Hence:


*[tex \huge xy = (12-y)y = 12y-y^2 = 163]


Put in standard form and solve:


*[tex \huge y^2 - 12y + 163 = 0 ]


*[tex \huge y =\frac {-(-12) \pm {sqrt{(-12)^2-4(1)(-163)}}} {2*1}=6 \pm i sqrt{127}]


Therefore:  *[tex \huge x = 12 - (6 \pm i sqrt{127}) = 6 \mp i sqrt{127}]


So:


*[tex \huge y = 6 + i sqrt{127} \text { } \Rightarrow \text { } \math x = 6 - i sqrt{127}]


and


*[tex \huge y = 6 - i sqrt{127} \text { } \Rightarrow \text { } \math x = 6 + i sqrt{127}]



Either way,


*[tex \huge x^2 + y^2 = (6 + i sqrt{127})^2 + (6 - i sqrt{127})^2]


*[tex \huge (6 + i sqrt{127})^2 + (6 - i sqrt{127})^2= (36 + 12 i sqrt(127) - 127) + (36 - 12 i sqrt{127} - 127) = 2( 36 - 127 ) = 2 * -91 = -182]