{x} , [x] & x are in GP find common ratio. and x is a +ve real no.

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Let x = I + f so that [x] = I>=0 and {x} = f.

So the equation can be written as I^2 = (I+f)f

Now $f = \frac{I^2}{ I+f} > \frac{I^2}{I+I} = \frac{I}{2}$

$\because 0\le f<1, \frac{I}{2}<1 \Rightarrow I<2 \Rightarrow I = 0,1$

I = 0 gives f=0

I =1 gives the quadratic f^2-f-1 = 0 which gives the permissible root $f = \frac{\sqrt 5-1}{2}$

Hence, there is a unique value of x among positive reals which is

$x = I+f = 1+ \frac{\sqrt 5-1}{2} = \frac{\sqrt 5+1}{2}$

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