Question

if x+y+z=6, and xy+yz+zx=9 then find the value of
$x {}^{2} + y {}^{2} + z { }^{2} =$


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Answer 1

Here , the formula is ,

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

and here given value is ,

x+y+z = 6 & xy+yz+xz = 9

from above formula ,

$(x + y + z) ^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz \\ \\ (x + y + z) ^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz ) \\ \\ (6) ^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(9) \\ \\ {x}^{2} + {y}^{2} + {z}^{2} + 18 = 36 \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = 36 - 18 \\ \\ {x}^{2} + {y}^{2} + {z}^{2} = 18$

Hope it helps....

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