Question

If x + y +z = 12, x2 + y2 + z2 = 64 , find the value of xy + yz + zx

Answer 1

Given,

x + y +z = 12, x² + y²+ z² = 64.

To find,

The value of xy+yz+zx.

Solution,

The expansion of (x + y + z)² = x²+y²+z²+2 (xy+yz+zx).

Putting the respective values in the given equation we get,

⇒   (x + y + z)² = x²+y²+z²+2 (xy+yz+zx).

⇒   (12)² = 64 + 2(xy+yz+zx)

⇒   144 = 64 + 2(xy+yz+zx)

64 in the right hand side of the equation would be shifted to left hand side of the equation and its sign would change from  =64 to -64.

⇒  144 - 64 = 2(xy+yz+zx)

⇒   80 = 2(xy+yz+zx)

⇒   xy+yz+zx = 80/2

⇒   xy+yz+zx= 40.

Hence, the value of xy+yz+zx = 40.