Solve the equation: sqrt(2x+3) - sqrt(x+1) = sqrt(x-2)

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

• $2x+3 \ge 0 \implies 2x \ge -3 \implies x \ge -\frac{3}{2}$
• $x+1 \ge 0 \implies x \ge -1$
• $x-2 \ge 0 \implies x \ge 2$ For all three conditions to be met, $x$ must be greater than or equal to 2. So, the domain is $x \ge 2$.

The given equation is $\sqrt{2x+3} - \sqrt{x+1} = \sqrt{x-2}$.
Move the negative radical term to the right side:

$$\sqrt{2x+3} = \sqrt{x-2} + \sqrt{x+1}$$

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

Subtract $2x-1$ from both sides:

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

Divide both sides by 2:

$$2 = \sqrt{x^2 - x - 2}$$

Square both sides:

$$(2)^2 = (\sqrt{x^2 - x - 2})^2$$ $$4 = x^2 - x - 2$$

$$x^2 - x - 2 - 4 = 0$$ $$x^2 - x - 6 = 0$$

Factor the quadratic expression:

$$(x-3)(x+2) = 0$$

This gives two potential solutions:

$$x-3 = 0 \implies x = 3$$ $$x+2 = 0 \implies x = -2$$

We must check these solutions against the domain $x \ge 2$.

• For $x=3$: This value satisfies $x \ge 2$. Substitute $x=3$ into the original equation:

$$\sqrt{2(3)+3} - \sqrt{3+1} = \sqrt{3-2}$$ $$\sqrt{9} - \sqrt{4} = \sqrt{1}$$ $$3 - 2 = 1$$ $$1 = 1$$

This solution is valid.

• For $x=-2$: This value does not satisfy $x \ge 2$. If we substitute $x=-2$ into $\sqrt{x-2}$, we get $\sqrt{-2-2} = \sqrt{-4}$, which is not a real number. Thus, $x=-2$ is an extraneous solution.

The only valid solution is $x=3$.