# How do you solve sqrt( 3x+1) - sqrt(x-1) = 2?

Oct 13, 2015

$x = 1$

#### Explanation:

You're dealing with radical terms, which means that you need to make sure that the expressions that are unde the square root are always positive.

This means that you need

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

$x - 1 \ge 0 \implies x \ge 1$

Combining these two conditions will get you $x \ge 1$. This means that any value of $x$ that is smaller than $1$ will cannot be a solution to the equation.

With this in mind, start by squaring both sides of the equation

${\left(\sqrt{3 x + 1} - \sqrt{x - 1}\right)}^{2} = {2}^{2}$

${\left(\sqrt{3 x + 1}\right)}^{2} - 2 \sqrt{\left(3 x + 1\right) \left(x - 1\right)} - {\left(\sqrt{x - 1}\right)}^{2} = 4$

$3 x + 1 - 2 \sqrt{\left(3 x + 1\right) \left(x - 1\right)} - x + 1 = 4$

$2 x - 2 \sqrt{\left(3 x + 1\right) \left(x - 1\right)} + 2 = 4$

Isolate the remaining square root on one side of the equation

$\sqrt{\left(3 x + 1\right) \left(x - 1\right)} = \frac{4 - 2 - 2 x}{2} = 1 - x$

Once again, square both sides of the equation

${\left(\sqrt{\left(3 x + 1\right) \left(x - 1\right)}\right)}^{2} = {\left(1 - x\right)}^{2}$

$\left(3 x + 1\right) \left(x - 1\right) = 1 - 2 x - {x}^{2}$

$3 {x}^{2} - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} - 1 = 1 - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} - {x}^{2}$

Rearrange the quadratic equation by moving all the terms on one side of the equation

$4 {x}^{2} - 2 = 0$

This is equivalent to

$2 \left({x}^{2} - 1\right) = 0 \implies {x}^{2} - 1 = 0 \iff \left\{\begin{matrix}x = - 1 \\ x = 1\end{matrix}\right.$

Since you need $x$ to satisfy the condition $x \ge 1$, only $x = 1$ will be a valid solution; $x = - 1$ will be an extraneous solution.

Do a quick check to make sure that the calculations are correct

$\sqrt{3 \cdot \left(1\right) + 1} - \sqrt{\left(1\right) - 1} = 2$

$\sqrt{4} - 0 = 2$

$2 = 2 \textcolor{w h i t e}{x} \textcolor{g r e e n}{\sqrt{}}$