Question

if 4x²+9y²+z²-6xy-3yz-2xz=0 then x:y:z is​

Answer 1

Answer:

Given,

x,y,z are real

and 

4x2+9y2+16z2−6xy−12yz−8zx=0

(2x)2+(3y)2+(4z)2−(2x)(3y)−(3y)(4z)−(2x)(4z)=0

[21× (2x−3y)2+(3y−4z)2+(4z−2x)2]=0

(2x−3y)2+(3y−4z)2+(4z−2x)2=0

sum of square can not be 0 so this  means these all are 0

2x=3y and 3y=4z

2x=3y=4z=k

so , x=2k , y=3k and z=4k

So, x,y,z are in H.P