Question

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

Answer 1

Answer:

x+y+z=9

xy+yz+xz=26

(x+y+z)² = x²+y²+z²+2xy+ 2yz+ 2xz

(9)²= x²+y²+z² + 2(xy+yz+xz)

81= x²+y²+z²+ 2(26)

81-52=x²+y²+z²

x²+y²+z²= 29

Answer 2

Answer = 29.

(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 )

= (9)^2 - 2(26)

= 81 - 52

= 29.