Question

X+y=3\4, xy= 2целых 5\8 .Найдиите x^2+y^2

Answer 1

$x+y=\frac{3}{4},\; \; xy=2\frac{5}{8}=\frac{21}{8}\\\\(x+y)^2=x^2+y^2+2xy\\\\(\frac{3}{4})^2=x^2+y^2+2\cdot \frac{21}{8}\\\\x^2+y^2=-\frac{9}{16}+\frac{21}{4}=\frac{-9+21\cdot 4}{16}=\frac{75}{16}=4,6875$