Question 372650
 
Sensible points : 
*) Definition of the set of x
*) Definition of Sqrt. (function or multi-valuate)

Supposition : x is in the set of real numbers, only real image allowed.
 

sqrt (5x-5) - sqrt (4x-3) = 1
 
Guessing : 1 is not solution (-1<>1), except if Sqrt(1)=+/- 1. 6 and 3 leaves not exact roots.
 
  => solution 1, if sqrt(1)=-1 (sqrt negatively defined)
 
 
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Problem by blind Equation manipulation : most probable steps 
 
1) root elimination : 
 1a) squaring
 1b) remaining root isolation
 1c) re-squaring -> 2nd degree equation
2) Verification 
 

1a) 5x-5 + 4x-3 -2Sqrt((5x-5)(4x-3)) = 1
1b) -2Sqrt((5x-5)(4x-3)) = 1 - 9x + 8
1c) 4(5x-5)(4x-3) = 9(1-x) = 20(x-1)(4x-3)
 
solution if x = 1 ? Not by the guessing point if sqrt is uni-valuate positive.
 
suppose x<>1, divide by x-1 : 
 
20(4x-3) = -9 = 80x - 60
 
80x = 51
 
x = 51/80
 
But x-1<0, hence 51/80 is not a solution since the roots are either real or imaginary.




 
Supposition 2 : x are complex numbers : x = a + ib
 
 
let sqrt(5x-5) = p+iq => 

1) p^2 - q^2 = 5a - 5
2) 2pq = 5b
 
sqrt(4x-3) = s + it =>
 
3) s^2 -t^2 = 4a - 3
4) 2st = 5b
 
and p-s = 1, q+t = 0

s=p-1, t = -q, get : (2)&(4) : 2(p-1)q = -4b = 2pq - 2q = 5b - 2q
 
q = 9b/2
 
then (2) : 2p9b/2 = 5b, if b<>0, then p = 5/9
 
equ (1) and (3) : 
 
5((p-1)^2-q^2+3) = 4(p^2 - q^2 + 5)
 
p^2 - q^2 = 10p
 
q^2 = p^2 - 10p < 0, hence there are no solution more in the complex numbers.
 

3rd possibility remains if x is a matrix.