Question

x+y+z=2, xy+yz+zx=1, (x+y)^2 +(y+z)^2 +( z+x)^2 =?

Answer 1

Answer:

6

Step-by-step explanation:

Given :

x + y + z = 2,

xy + yz + zx = 1,

Then, (x + y)² + (y + z)² + (z + x)² = ?

Solution :

We know that,

x + y + z = 2

⇒ (x + y + z) = (2)²

⇒ x² + y² + z² + 2xy + 2yz + 2zx = 4

⇒ x² + y² + z² + 2(xy + yz + zx) = 4

⇒ x² + y² + z² + 2(1) = 4

⇒ x² + y² + z² + 2 = 4

⇒  x² + y² + z² = 4 - 2 = 2

⇒ x² + y² + z² = 2 ...(i)

Now,

(x + y)² + (y + z)² + (z + x)²

⇒ (x² + 2xy + y²) + (y² + 2yz + z²) + (z² + 2zx + x²)

⇒ 2x² + 2y² + 2z² + 2xy + 2yz + 2zx

⇒ 2(x² + y² + z²) + 2(xy + yz + zx)

⇒ 2(2) + 2(1) = 4 + 2 = 6