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

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Answer 1 · 2018-03-25T17:16:58

Answer is (a).

#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 · 2018-03-25T17:31:37

#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#