Question

tf (x+y)=6 and xy=4 then find the value of x^2+y^2 .

Answer 1

(x+y)2= x2+y2+2xy

(6)2=x2+y2+2×4

36-8=28= x2+y2

Answer 2

${x}^{2} + {y}^{2} \\ {(x + y)}^{2} - 2xy \\ {6}^{2} - 2 \times 4 \\ 36 - 8 \\ 28$