The given equation is:

$\sqrt{2 + x \sqrt{2x + 3} - 3} = 0$

Let's solve this step by step:

Step 1: Set the square root term equal to zero.

$\sqrt{2 + x \sqrt{2x + 3} - 3} = 0$

To remove the square root, square both sides of the equation: $2 + x \sqrt{2x + 3} - 3 = 0^2$ $2 + x \sqrt{2x + 3} - 3 = 0$ Simplify: $x \sqrt{2x + 3} - 1 = 0$

Step 2: Isolate the square root term.

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

Step 3: Square both sides again to eliminate the square root.

$x^2 (2x + 3) = 1^2$ $x^2 (2x + 3) = 1$

Step 4: Expand and simplify.

$x^2(2x) + x^2(3) = 1$ $2x^3 + 3x^2 = 1$

Step 5: Rearrange the equation.

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

This is a cubic equation that needs solving. Now, we can test the options given in the multiple-choice (3, 23, and so on) to see which one satisfies the equation.

Let's test each option:

1. Option (a): $x = 3$

   Substituting $x = 3$ into the cubic equation: $2(3)^3 + 3(3)^2 - 1 = 2(27) + 3(9) - 1 = 54 + 27 - 1 = 80 \neq 0$ So, $x = 3$ is not a solution.

2. Option (b): $x = 23$

   Substituting $x = 23$ into the cubic equation: $2(23)^3 + 3(23)^2 - 1 = 2(12167) + 3(529) - 1 = 24334 + 1587 - 1 = 25920 \neq 0$ So, $x = 23$ is not a solution.

3. Option (c): $x = 3 \text{ and } 23$

   Since neither $x = 3$ nor $x = 23$ satisfy the equation, this option is also incorrect.

4. Option (d): $x = 20$

   Substituting $x = 20$ into the cubic equation: $2(20)^3 + 3(20)^2 - 1 = 2(8000) + 3(400) - 1 = 16000 + 1200 - 1 = 17200 - 1 = 17199 \neq 0$ So, $x = 20$ is not a solution either.

Thus, none of the options seem to directly solve the equation based on typical algebraic methods. If needed, we could attempt further numerical techniques or a graphing approach to find the solution.

Would you like more details or have any questions?

Here are 5 related questions that expand on this topic:

1. How do we solve a cubic equation using numerical methods like the Newton-Raphson method?
2. What are the conditions under which squaring both sides of an equation introduces extraneous solutions?
3. Can all cubic equations be factored, and if so, how can we recognize when they can be?
4. What is the best approach to solving complex radical equations?
5. How does graphing help in finding the solutions to cubic equations?

Tip: Always check for extraneous solutions when solving equations that involve squaring both sides.