What is true about the solution of \( \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} ? \) \( x= \pm \sqrt{3} \), and they are actual solutions. \( x= \pm \sqrt{3} \), but they are extraneous solutions. \( x=3 \), and it is an actual solution. \( x=-3 \), but it is an extraneous solution.

To solve the equation \( \frac{x^{2}}{2 x-6}=\frac{9}{6 x-18} \), you first notice that both sides can be simplified, leading to the identification of critical points. Remember, while solving such equations, it’s vital to check for restrictions on the variable due to division by zero. When checking, you'll find that the simplifications yield \( x = 3 \) as a potential solution, but it also turns out to be a point where the denominator becomes zero. Therefore, it’s considered an extraneous solution. Always check back with the original equation to confirm if solutions are valid or simply illusions created through the solving process!