Question

If x + y = 6, y + z = 11 and x + z = 9 and xy + yz + zx = 50 then find the value of x² +y² + z².​

Answer 1

$\begin{gathered}x + y + z = 6 \\ = > on \: squaring \: both \: sides \\ = > (x + y + z) ^{2} = (6) ^{2} \\ = > x ^{2} + y ^{2} + z ^{2} + 2xy + \\ 2yz + 2zx = 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 2(xy + \\ yz + zx) = 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 2 \times 11 = \\ 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 22 = 36 \\ = > x ^{2} + y ^{2} + z ^{2} = 36 - 22 \\ = > x ^{2} + y ^{2} + z ^{2} = 14\end{gathered}$

Answer 2

x+y+z)=6square in both sides,(x+y+z)^2=6^2×^2+y