Solve the equation: sqrt(3x+4) − sqrt(x+5) = 1

For the square roots to be defined, the expressions under them must be non-negative:

$$3x+4 \ge 0 \implies 3x \ge -4 \implies x \ge -\frac{4}{3}$$ $$x+5 \ge 0 \implies x \ge -5$$

Combining these conditions, we must have $x \ge -\frac{4}{3}$.

The given equation is $\sqrt{3x+4} - \sqrt{x+5} = 1$.
Move the second radical term to the right side of the equation:

$$\sqrt{3x+4} = 1 + \sqrt{x+5}$$

$$(\sqrt{3x+4})^2 = (1 + \sqrt{x+5})^2$$ $$3x+4 = 1^2 + 2(1)\sqrt{x+5} + (\sqrt{x+5})^2$$ $$3x+4 = 1 + 2\sqrt{x+5} + x+5$$ $$3x+4 = x+6 + 2\sqrt{x+5}$$

Subtract $x+6$ from both sides:

$$3x+4 - x - 6 = 2\sqrt{x+5}$$ $$2x - 2 = 2\sqrt{x+5}$$

Divide both sides by 2:

$$x - 1 = \sqrt{x+5}$$

$$(x-1)^2 = (\sqrt{x+5})^2$$ $$x^2 - 2x + 1 = x+5$$

$$x^2 - 2x - x + 1 - 5 = 0$$ $$x^2 - 3x - 4 = 0$$

Factor the quadratic equation:

$$(x-4)(x+1) = 0$$

This gives two possible solutions:

$$x_1 = 4$$ $$x_2 = -1$$

Recall the domain $x \ge -\frac{4}{3}$. Both $x=4$ and $x=-1$ satisfy this condition.
Substitute each solution into the original equation $\sqrt{3x+4} - \sqrt{x+5} = 1$:

For $x=4$:

$$\sqrt{3(4)+4} - \sqrt{4+5} = \sqrt{12+4} - \sqrt{9}$$ $$= \sqrt{16} - 3 = 4 - 3 = 1$$

This matches the right side of the equation, so $x=4$ is a valid solution.

For $x=-1$:

$$\sqrt{3(-1)+4} - \sqrt{-1+5} = \sqrt{-3+4} - \sqrt{4}$$ $$= \sqrt{1} - 2 = 1 - 2 = -1$$

This does not match the right side of the equation (which is 1), so $x=-1$ is an extraneous solution.

The only valid solution is $x=4$.