Simplest way to solve radical equation $\sqrt{3x+1}-\sqrt{x+4}=1$ I have the following equation:
$$\sqrt{3x+1}-\sqrt{x+4}=1$$
I can get the answer $x=5$ through tedious and long algebraic manipulation with quite a few extraneous solutions. It's not elegant. Is there a simple, straightforward way to solve this equation?
 A: The domain is $x\geq-\frac{1}{3}$ and squaring:
$$3x+1=1+2\sqrt{x+4}+x+4$$ or
$$\sqrt{x+4}=x-2,$$
which gives also $x\geq2$.
$$x+4=(x-2)^2$$ or
$$x(x-5)=0,$$ which gives the answer:
$$\{5\}$$
A: Condition: $x\geq -\frac{1}{3}$.
One has $\sqrt{3x+1} = \sqrt{x+4} + 1$, then $3x+1 = x+4 +1 + 2\sqrt{x+4}$.
So $x-2 = \sqrt{x+4}$. Note that $x \geq 0$.
Then $x^2-4x+4=x+4$, or $x^2-5x = 0$. So $x=5$.
A: What do you consider long and tedious.
I usually isolate one radical to one side by itself and square, repeat, keep track of signage and it usually works well
$\sqrt{3x+1}-\sqrt{x+4}=1$
$\sqrt{3x + 1} = 1 + \sqrt{x+4}$ [make note $3x+1 \ge 1; x+4 \ge 0$]
$3x + 1 = (1 + \sqrt{x+4})^2 = 1 + 2\sqrt{x+4} + x+4$
$2x -4 = 2\sqrt{x+4}$
$x -2 = \sqrt{x+4}$ [Make note $x-2 \ge 0$
$(x-2)^2 = x + 4$
$x^2 - 4x +4 = x + 4$
$x^2 -5x = 0$.
Quadratic equation or in this case simple factoring yields:
$x(x-5)= 0$
So $x = 0$ or $x = 5$.  As $0 - 2 < 0$ so that's out.  So $x = 5$.
A: If we introduce the variables 
\begin{align}
y^2&=&3x+1\tag1\\
z^2&=&x+4\tag2\\
y-z&=&1\tag3\end{align}
we get an extraneous solution, but it is easy to dismiss and this makes the calculations a bit neater.  Subtract three times $(2)$ from $(1)$
\begin{align}
y^2-3z^2=-11\\\
\end{align}
using $(3)$
\begin{align}
(z+1)^2-3z^2+11=-2z^2+2z+12=0\\
\end{align}
$$\Leftrightarrow$$
\begin{align}
z^2-z-6=0\\
\end{align}
There are two solutions to this $z = -2$ and $z = 3$. We dismiss the first since $z$ represents a square root and we don't want this to be negative. The second leads to $x=5$, $y=4$ and $z=3$.
