|
(I) Applying Componendo and Dividendo, we have:
x+z)/y = z/x = x/(z-y) = (x+z+z+x)/(x+z) = 2
or, x+z =2y....(i)
z=2x......(ii)
x=2(z-y)...(iii)
Then, from (i) and(ii), we have:
3x= 2y
=> y= 3x/2
Also, x+z=2y
or, z= 2y -x = 3x-x = 2x
Also, 2(z-y) = 2(2x -3x/2) = x, so that:
So,(y,z) = (3x/2, 2x) also satisfies (iii)
Thus, x:y:z= 2:3:4
(II) 21(x^2+y^2+z^2) = (x+2y+4z)^2
or, 20x^2 + 17y^2 + 5z^2 - 4xy - 8xz - 16yz =0
or,(2x-y)^2 + (4x-z)^2 + (4y-2z)^2 = 0
Since each of x,y, and z is a real number, this is possible only when:
2x=y, 4x=z, and 4y=2z
or, x:y =1:2, x:z = 1:4, and: y:z=. 2:4
or, x:y:z = 1:2:4
|
