We want to solve the equation:

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

Steps to solve:

1. Square both sides to eliminate the square root: $(\sqrt{3 - 2x})^2 = x^2$ $3 - 2x = x^2$

2. Rearrange into standard quadratic form: $x^2 + 2x - 3 = 0$

3. Factorize the quadratic: $(x + 3)(x - 1) = 0$

4. Solve for $x$: $x = -3 \quad \text{or} \quad x = 1$

5. Check for extraneous solutions: Substitute each solution back into the original equation:

   • For $x = -3$: $\sqrt{3 - 2(-3)} = \sqrt{3 + 6} = \sqrt{9} = 3 \neq -3$ $x = -3$ is not valid.
   • For $x = 1$: $\sqrt{3 - 2(1)} = \sqrt{3 - 2} = \sqrt{1} = 1$ $x = 1$ is valid.

Final Solution:

$x = 1$

Would you like further details or have any questions?

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Here are five related questions to expand on this topic:

1. What are extraneous solutions, and why do they occur in equations involving square roots?
2. How do you verify solutions to equations involving square roots?
3. Can this equation have solutions that aren't real numbers? Why or why not?
4. How would the solution change if the equation were $\sqrt{3 - 2x} = -x$?
5. What methods are used to solve more complex radical equations?

Tip:

Always check your solutions in the original equation when working with radicals or exponents to ensure they satisfy the equation.