How do you solve #1+sqrt(3x+1)=x# and find any extraneous solutions?

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Answer 1 · 2016-06-09T11:20:25

First, the argument under the root must be non-negative #(>=0)#

#3x+1>=0->x>=-1/3#

Subtract #1# from both sides:
#sqrt(3x+1)=x-1#

Now we square both sides:
#(sqrt(3x+1))^2=(x-1)^2->#

#3x+1=x^2-2x+1->#

This can be normalised to:
#x^2-5x=0->x(x-5)=0->#

#x=0orx=5#

The first solution is false, as #1+sqrt(3xx0+1)!=0#
The other one gives #1+sqrt(3xx5+1)=5#

Conclusion: #x=5#