Question

if x^2+y^2+z^2=78 and xy+yz+zx=59. Find x+y+z​

Answer 1

Answer:

by identity

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

(x+y+z)^2=78+2*59

x+y+z=14

Step-by-step explanation:

Answer 2

Step-by-step explanation:

(X + y + z )²=X ² + y ²+ z ² + 2 X y + 2 y z + 2 x z

(X + y + z)² =78 +2(x y +y z +z x)

= 78+2(59)

=78+118

=196

(X+y+z)=√196

X+y+z = 14 ans