Simplest way to solve radical equation $\sqrt{3x+1}-\sqrt{x+4}=1$

Question

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?

Answer 1

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: $$\{5\}$$

Answer 2

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 $x=5$.

Answer 3

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 $x = 5$.

Answer 4

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 $z=3$.