Let's solve the equation:

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

Step 1: Isolate one of the square roots

To start, let's isolate one of the square roots. We can move $\sqrt{x}$ to the other side of the equation:

$\sqrt{3x+4} = \sqrt{x} + 2$

Step 2: Square both sides

Now, we'll square both sides of the equation to eliminate the square root on the left-hand side:

$\left( \sqrt{3x+4} \right)^2 = \left( \sqrt{x} + 2 \right)^2$

This simplifies to:

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

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

Step 3: Simplify the equation

Now, simplify both sides:

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

Subtract $x + 4$ from both sides:

$2x = 4\sqrt{x}$

Step 4: Isolate $\sqrt{x}$

Next, divide both sides of the equation by 4:

$\frac{2x}{4} = \sqrt{x}$

Simplifying:

$\frac{x}{2} = \sqrt{x}$

Step 5: Square both sides again

Now square both sides again to eliminate the remaining square root:

$\left( \frac{x}{2} \right)^2 = (\sqrt{x})^2$

This simplifies to:

$\frac{x^2}{4} = x$

Step 6: Solve the quadratic equation

Multiply both sides by 4 to eliminate the denominator:

$x^2 = 4x$

Now rearrange the terms:

$x^2 - 4x = 0$

Factor the quadratic:

$x(x - 4) = 0$

Step 7: Solve for $x$

This gives two possible solutions:

$x = 0 \quad \text{or} \quad x = 4$

Step 8: Check for extraneous solutions

We must check both solutions in the original equation because squaring both sides can introduce extraneous solutions.

For $x = 0$:

Substitute $x = 0$ into the original equation:

$\sqrt{3(0) + 4} - \sqrt{0} = 2$

$\sqrt{4} - 0 = 2$

$2 = 2$

So, $x = 0$ is a valid solution.

For $x = 4$:

Substitute $x = 4$ into the original equation:

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

$\sqrt{12 + 4} - 2 = 2$

$\sqrt{16} - 2 = 2$

$4 - 2 = 2$

So, $x = 4$ is also a valid solution.

Final Answer:

The solutions to the equation are:

$x = 0 \quad \text{or} \quad x = 4$

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Would you like more details or have any questions? Here are a few related questions for you:

1. How can we solve equations involving square roots when there are more than two terms?
2. What is the effect of squaring both sides of an equation?
3. How can we check for extraneous solutions in equations with square roots?
4. What are the steps involved in solving a quadratic equation?
5. How do you solve equations where the variable is inside a square root?

Tip: Always check for extraneous solutions after squaring both sides of an equation, as squaring can introduce solutions that do not satisfy the original equation.