Question

If x + y=6 , y + z=11 and x + z=9 and xy + yz + zx=50 then find the value of x2 +y2 + z2.​

Answer 1

Answer:

(x + y) + (y + z) + (x + z) =11+6+9

2(x + y + z) = 26

2x +2y +2z =26

Answer 2

Answer :

If x + y = 6, y + z = 11 and x + z = 9 and xy + yz + zx = 50, then the value of x² + y² + z² is 69.

Step-by-step explaination :

According to the question :

➟ x + y = 6 . . . . . ⑴

➟ y + z = 11 . . . . . ⑵

➟ x + z = 9 . . . . . ⑶

➟ xy + yz + zx = 50

We shall find the value of :

➟ x² + y² + z²

Now, let’s add ⑴, ⑵ and ⑶ to get x + y + z :

➟ (x + y) + (y + z) + (z + x) = 6 + 11 + 9

➟ x + x + y + y + z + z = 26

➟ 2x + 2y + 2z = 26

➟ 2 (x + y + z) = 26

➟ x + y + z = 26/2

➟ x + y + z = 13

We know the algebraic identity :

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

Let’s substitute the values given and find x² + y² + z² :

➟ (13)² = x² + y² + z² + 2 (50)

➟ 169 = x² + y² + z² + 100

➟ 169 – 100 = x² + y² + z²

➟ 69 = x² + y² + z²

Therefore, the value of x² + y² + z² is 69.