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

Jun 9, 2016

First, the argument under the root must be non-negative $\left(\ge 0\right)$

#### Explanation:

$3 x + 1 \ge 0 \to x \ge - \frac{1}{3}$

Subtract $1$ from both sides:
$\sqrt{3 x + 1} = x - 1$

Now we square both sides:
${\left(\sqrt{3 x + 1}\right)}^{2} = {\left(x - 1\right)}^{2} \to$

$3 x + 1 = {x}^{2} - 2 x + 1 \to$

This can be normalised to:
${x}^{2} - 5 x = 0 \to x \left(x - 5\right) = 0 \to$

$x = 0 \mathmr{and} x = 5$

The first solution is false, as $1 + \sqrt{3 \times 0 + 1} \ne 0$
The other one gives $1 + \sqrt{3 \times 5 + 1} = 5$

Conclusion: $x = 5$