Question

If x+y+z=9, xy+yz+zx =26; find x²+y²+z²

Answer 1

we know,
(x + y + z)² = x² + y² + z² + 2(xy + yz + xz)
9² = x² + y² + z² + 2 X 26
81 = x² + y² + z² + 52
81 - 52 = x² + y² + z²
therefore, x² + y² + z² = 29

Answer 2

First, we must expand and simplify :

`x + y+z = 9`


`(x + y+z)^(2) = (9)^(2)`


`x^(2) + y^(2)+z^(2) + 2(xy + yz+zx) = 81`

And Than, Enter the value :

`x^(2) + y^(2)+z^(2) + 2(26) = 81`


`x^(2) + y^(2)+z^(2) + 52 = 81`


`x^(2) + y^(2)+z^(2) = 81- 52`


`\boxed{x^(2) + y^(2)+z^(2) = 29}`