Question

Let `x,y,z` are three real and distinct numbers which satisfy the Equation `8(4x^2+y^2)+2z^2-4(4xy+yz+2xz)=0`,then Which of The following options are correct? (a) `x/y=1/2` (b)`y/z=1/4` (c)`x/y=1/3` (d)x,y,z are in A.P

Answer 1

Answer is (a).

Explanation:

`8(4x^2+y^2)+2z^2-4(4xy+yz+2xz)=0` can be written as

`32x^2+8y^2+2z^2-16xy-4yz-8xz=0`

or `16x^2+4y^2+z^2-8xy-2yz-4xz=0`

i.e. `(4x)^2+(2y)^2+z^2-4x*2y-2y*z-4x*z=0`

if `a=4x`, `b=2y` and `c=z`, then this is

`a^2+b^2+c^2-ab-bc-ca=0`

or `2a^2+2b^2+2c^2-2ab-2bc-2ca=0`

or `(a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ac)=0`

or `(a-b)^2+(b-c)^2+(c-a)^2=0`

Now if sum of three squares is `0`, they must each be zero.

Hence `a-b=0`, `b-c=0` and `c-a=0`

i.e. `a=b=c` and in our case `4x=2y=z=k` say

then `x=k/4`, `y=k/2` and `z=k`

i.e. `x,y` and `z` are in G.P, and `x/y=2/4=1/2`

`y/z=1/2` and hence answer is (a).

Answer 2

`x,y,z` are three real and distinct numbers which satisfy the Equation
Given

`8(4x^2+y^2)+2z^2-4(4xy+yz+2xz)=0`

`=>32x^2+8y^2+2z^2-16xy-4yz-8xz=0`

`=>16x^2+4y^2-16xy +16x^2+z^2-8xz+4y^2+z^2-4yz=0`

`=>[(4x)^2+(2y)^2-2*4x*2y]+[(4x)^2+z^2-2*4x*z]+[(2y)^2+z^2-2*2y*z]=0`

`=>(4x-2y)^2+(4x-z)^2+(2y-z)^2=0`

Sum three squared real quantities being zero each of them must be zero.
Hence `4x-2y=0->x/y=2/4=1/2to`Option (a)

`4x-z=0=>4x=z`

and

`2y-z=0=>2y=z`