Question

If x+y+z=6 and xy+yz+zx=12 then find the value of x2+y2+z2

Answer 1

$xy + yz + zx = 12 \\ \\ x + y + z = 6 \\ \\ squaring \: both \: sides \\ \\ {(x + y + z)}^{2} = {(6)}^{2} \\ {x }^{2} + {y}^{2} + { z }^{2} + 2xy + 2yz + 2zx = 36 \\ {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) = 36 \\ {x}^{2} + {y}^{2} + {z}^{2} + 2(12) = 36 \\ {x}^{2} + {y}^{2} + {z}^{2} + 24 = 36 \\ {x}^{2} + {y}^{2} + {z}^{2} = 36 - 24 \\ {x}^{2} + {y}^{2} + {z}^{2} = 12$