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

Jun 6, 2016

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 $\sqrt{3 x + 1}$ to exist, we have to have a non-negative expression under a square root:
$3 x + 1 \ge 0$, that is $x \ge - \frac{1}{3}$

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

Combining the two conditions above, $x \ge - \frac{1}{3}$ and $x \ge 1$, we come up with one condition that defines the domain where the solution should be found:
$x \ge 1$

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

The solution $x = 5$ fits the restriction $x \ge 1$, 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: $\sqrt{3 \cdot 5 + 1} = \sqrt{16} = 4$
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.