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#