How would yous solve $sqrt(3x+1)-1=sqrt(8x-1)$ ?

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Answer 1 · 2018-05-15T15:01:49

$x=0.129$

In the equation $sqrt(3x+1)-1=sqrt(8x-1)$ we can only have $3x+1>=0$ and $8x-1>=0$ i.e. $x>=-1/3$ and $x>=1/8$ i.e. $x>=1/8$ or $x>=0.125$ .

To solve the equation $sqrt(3x+1)-1=sqrt(8x-1)$

squaring each side, we get

$3x+1-2sqrt(3x+1)+1=8x-1$

now take irrational portion on the left and we get

$-2sqrt(3x+1)=8x-1-3x-1-1=5x-3$

squaring again $4(3x+1)=25x^2-30x+9$

or $25x^2-30x+9=12x+4$

or $25x^2-42x+5=0$

or $x=(42+-sqrt(42^2-500))/50=(42+-sqrt1264)/50=(42+-35.553)/50$

i.e. $x=0.129$ or $1.551$

However on checking $1.551$ is not found to be a solution and hence only answer is $x=0.129$

How would yous solve sqrt(3x+1)-1=sqrt(8x-1)sqrt(3x+1)-1=sqrt(8x-1)?