Question

if x+y+z=9 and xy+yz+zx=26 find x^2+y^2+z^2.

Answer 1

given:

x+y+z=9

xy+yz+zx=26



Using Identity (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab+2bc + 2ca

so,


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


substitute for x+y+z and xy+yz+zx

and you'll get

(9)^2 = x^2 + y^2 + z^2 + 2 (26)
81 = x^2 + y^2 + z^2 + 52
81 - 52 = x^2 + y^2 + z^2
29 = x^2 + y^2 + z^2

so,
x^2 + y^2 + z^2 = 29

Answer 2

the picture given is the answer