SOLUTION: What is the solution for this square root equation? sqrt (5x-5) - sqrt (4x-3 =1) I am not sure how to solve this problem as i keep coming up with multiple answers. Please Hel

Question 372650: What is the solution for this square root equation?
sqrt (5x-5) - sqrt (4x-3 =1)
I am not sure how to solve this problem as i keep coming up with multiple answers. Please Help. Thank you.
Found 2 solutions by Fombitz, Jk22:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

Square both sides.

Square both sides again.

Two solutions:

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.
.

.
.
.
Verify both solutions by plugging them back into the original equation.
Remember .

Answer by Jk22(389) (Show Source): You can put this solution on YOUR website!

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.

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