Question

If x +y+z=9 and xy+yz+xz=26 find the value of x^2+y^2+z^2

Answer 1

Answer:

Step-by-step explanation:

X+y+z=9

xy+yz+zx=26

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

        81       =(x^2+y^2+z^2)+ 2*26

             x^2+y^2+z^2=81-54

                                  = 27

                =63

Answer 2

Answer:

55

Step-by-step explanation:

(x +y+z)^2=81

Or, x^2+y^2+z^2 + 2(xy+yz+xz) = 81

Or, x^2+y^2+z^2 = 81 - 26 = 55