How do you solve #sqrt(3x+1) = x1# and find any extraneous solutions?
1 Answer
Answer:
The only solution:
An extraneous "solution" acquired by noninvariant transformation of the equation (raising to the power of
Explanation:
We have to start by specifying the domain where the solutions can be found.

For
#sqrt(3x+1)# to exist, we have to have a nonnegative expression under a square root:
#3x+1>=0# , that is#x>=1/3# 
When we use just
#sqrt(A)# (not#+sqrt(A)# ), we assume the nonnegative value, the square of which equals to#A# (and#A>=0# as mentioned above). Therefore, the right side of the given equation must be nonnegative:
#x1 >= 0# , that is#x>=1#
Combining the two conditions above,
Now let's simplify the equation by raising to power of
This equation has two solutions:
The solution
So, the only solution seems to be
CHECK
Left side:
Right side:
So, the equation is satisfied with
We recommend to study a series of lectures on how to solve equations at Unizor by following the menu options Algebra  Equations and Inequalities.