Question

if x+y+z=8 and xy+yz+zx=18, find the value of x²+y²+z²​

Answer 1

${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx)$

${x}^{2} + {y}^{2} + {z}^{2} = {(x + y + z)}^{2} - 2(xy + yz + zx)$

${x}^{2} + {y}^{2} + {z}^{2} = {8}^{2} - (2 \times 18)$

${x}^{2} + {y}^{2} + {z}^{2} = 64 - 36$

${x}^{2} + {y}^{2} + {z}^{2} = 28$

Hope it helps you!

Answer 2

Answer:

find the value of x²+y²+z² if x+y+z=18 and xy+yz+xx=46