Question

If x2 + 4y2 + z2 - 2xy - 2yz - zx = 0 then x:y: z equals
(A) 1:2:1
(B) 2:1:2
(C) 1:2:3​

Answer 1

Answer:

A) 1:2:1

Is your answer mate

Answer 2

Answer:

(B) 2:1:2

Step-by-step explanation:

Given, $x^2 +4y^2+ z^2-2xy-2yz-zx = 0$....(1)

We now multiply equation (1) with 2

Equation(1) becomes

$2x^2 +2(2y)^2+ 2z^2-4xy-4yz-2zx = 0$

⇒ $[{x^2+(2y)^2-2(2xy)]+[x^2+z^2-2xz]+[z^2+(2y)^2-2(2zy)] =0$(Rearranging the terms)

⇒$[(x-2y)^2]+[(x-z)^2]+[(z-2y)^2] =0$

This is only possible if $x- 2y=0$, $x-z =0$, and $z-2y = 0$.....(2)

Now, let $x=k$

  then using equations (2) we get:

               $z=k\\ y = k/2$

Therefore, $x:y:z = k:k/2:k$

i.e. $x:y:z = 2:1:2$

#SPJ2