The equation in the image is:

$\sqrt{3 - x} - \sqrt{7 + x} = \sqrt{16 + 2x}$

Let's solve this step by step.

Step 1: Isolate one of the square root terms

Move $\sqrt{7 + x}$ to the right-hand side:

$\sqrt{3 - x} = \sqrt{16 + 2x} + \sqrt{7 + x}$

Step 2: Square both sides to eliminate the square root

Squaring both sides will help us remove the square root on the left side.

$(\sqrt{3 - x})^2 = (\sqrt{16 + 2x} + \sqrt{7 + x})^2$

This simplifies to:

$3 - x = (16 + 2x) + (7 + x) + 2\sqrt{(16 + 2x)(7 + x)}$

Simplify the equation:

$3 - x = 23 + 3x + 2\sqrt{(16 + 2x)(7 + x)}$

Step 3: Isolate the square root again

Subtract $23 + 3x$ from both sides:

$-20 - 4x = 2\sqrt{(16 + 2x)(7 + x)}$

Step 4: Square both sides again

Square both sides to eliminate the square root:

$(-20 - 4x)^2 = 4[(16 + 2x)(7 + x)]$

Step 5: Expand and simplify

Expand both sides:

$400 + 160x + 16x^2 = 4[(16)(7) + 16x + 14x + 2x^2]$

Simplify:

$400 + 160x + 16x^2 = 4[112 + 30x + 2x^2]$

Further simplifying:

$400 + 160x + 16x^2 = 448 + 120x + 8x^2$

Step 6: Bring all terms to one side

$16x^2 - 8x^2 + 160x - 120x + 400 - 448 = 0$

$8x^2 + 40x - 48 = 0$

Step 7: Solve the quadratic equation

Divide the entire equation by 8:

$x^2 + 5x - 6 = 0$

Factorize the quadratic equation:

$(x + 6)(x - 1) = 0$

So, the solutions for $x$ are:

$x = -6 \quad \text{or} \quad x = 1$

Step 8: Check the solutions in the original equation

1. For $x = -6$:

$\sqrt{3 - (-6)} - \sqrt{7 + (-6)} = \sqrt{16 + 2(-6)}$

Simplifies to:

$\sqrt{9} - \sqrt{1} = \sqrt{4}$

$3 - 1 = 2 \quad \text{(True)}$

2. For $x = 1$:

$\sqrt{3 - 1} - \sqrt{7 + 1} = \sqrt{16 + 2(1)}$

Simplifies to:

$\sqrt{2} - \sqrt{8} \neq \sqrt{18}$

This is not true.

Final Answer:

The valid solution is $x = -6$.

Would you like more details or have any questions? Here are some related questions for further practice:

1. How do you square both sides of an equation correctly?
2. What is the importance of checking solutions in the original equation?
3. How do you factorize a quadratic equation?
4. What are the common mistakes to avoid when solving equations involving square roots?
5. How does squaring both sides affect the solutions?

Tip: When dealing with square roots, always check your solutions in the original equation to avoid extraneous solutions.