How do I solve this???

label Mathematics
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Nov 5th, 2014

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

Nov 5th, 2014

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