F(x) = 2 Solutions

Solution 1: We are given that $f(x) = 2 = \sqrt{3x+1} - \sqrt{x-1}$ . We proceed as follows:

$\begin{aligned} \sqrt{3x+1} &= 2+\sqrt{x-1} \\ 3x+1 &= 4 + 4\sqrt{x-1}+x-1 &\mbox{Squared both sides} \\ 0 & = 2-2x+4\sqrt{x-1} \\ 0 & = 1 - x + 2\sqrt{x-1} \\ x - 1 & = 2\sqrt{x-1} \\ x^2 - 2x + 1 & = 4(x-1) &\mbox{Squared both sides}\\ 0 &= x^2 - 6x + 5 \\ 0 &= (x-5)(x-1) \\ \end{aligned}$

Thus $x = 5$ and $x =1$ are possible solutions. We check that $f(5) = \sqrt{3(5)+1} - \sqrt{5-1} = 4 -2 = 2$ and $f(1) = \sqrt{3(1)+1} - \sqrt{1-1} = 2 - 0 = 2$ give valid values. Hence the sum of the possible values of $x$ is $1 + 5 = 6$ .

Solution 2: We want to find solutions to $2 = \sqrt{3x+1} - \sqrt{x-1}$ . For these values of $x$ , we have $2x+2=(3x+1)-(x-1)=(\sqrt{3x+1}+\sqrt{x-1} )(\sqrt{3x+1}-\sqrt{x-1} )$ $\Rightarrow x+1 = \sqrt{3x+1} + \sqrt{x-1}$ . Thus, $\sqrt{3x+1} = \frac {2 + x+1} {2}$ . Squaring both sides, we obtain $12x + 4 = x^2 + 6x + 9$ , or that $0=x^2-6x+5=(x-5)(x-1)$ . We can quickly check that 5 and 1 are solutions to the original equation. Hence, the sum of all the possible values of $x$ is $1 + 5 = 6$ .

Note: It is essential to check the solutions we found, since we may have introduced more solutions when squaring an equation.