squaring both sides

[sqrt(3x) +1]^2=[sqrt(5x+1)]^2

using (a + b)^2= a^2 + b^2 + 2ab, we get

3x+1+2sqrt(3x) = 5x+1

transfer (3x+1) to right hand side, we get

2 sqrt(3x) = 5x+1 - 3x -1

or, 2sqrt(3x) =2x

or, sqrt(3x) =x

again squaring both sides, we get

[sqrt(3x)]^2 = x^2

3x = x^2

transfer 3x to right hand side

we get

x^2 - 3x =0

take x common

x(x-3)=0

either x = 0 or, x-3= 0

x = 0 or, x=3

Answer : x=0 , 3

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