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

1 Answer
Jun 6, 2016

Answer:

The only solution: #x=5#
An extraneous "solution" acquired by non-invariant transformation of the equation (raising to the power of #2#) is #x=0# - THIS IS NOT A SOLUTION.

Explanation:

We have to start by specifying the domain where the solutions can be found.

  1. For #sqrt(3x+1)# to exist, we have to have a non-negative expression under a square root:
    #3x+1>=0#, that is #x>=-1/3#

  2. When we use just #sqrt(A)# (not #+-sqrt(A)#), we assume the non-negative value, the square of which equals to #A# (and #A>=0# as mentioned above). Therefore, the right side of the given equation must be non-negative:
    #x-1 >= 0#, that is #x>=1#

Combining the two conditions above, #x>=-1/3# and #x>=1#, we come up with one condition that defines the domain where the solution should be found:
#x>=1#

Now let's simplify the equation by raising to power of #2# both sides:
#3x+1 = (x-1)^2# or, simplifying,
#x^2-5x=0#
This equation has two solutions: #x=0# and #x=5#.

The solution #x=5# fits the restriction #x>=1#, but the "solution" #x=0# does not.
So, the only solution seems to be #x=5#, while #x=0# is an extraneous solutions (NOT THE REAL SOLUTION) that we acquired by raising the original equation to the power of #2# (a non-invariant transformation).

CHECK
Left side: #sqrt(3*5+1) = sqrt(16) = 4#
Right side: #5-1 = 4#
So, the equation is satisfied with #x=5#.

We recommend to study a series of lectures on how to solve equations at Unizor by following the menu options Algebra - Equations and Inequalities.