Question

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

Answer 1

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.