1. The problem is to find the extraneous root of an equation, which typically arises when solving equations involving radicals or rational expressions. 2. An extraneous root is a solution that emerges from the algebraic process but does not satisfy the original equation. 3. To identify extraneous roots, solve the equation normally, then substitute each solution back into the original equation to verify if it holds true. 4. Since the user did not provide a specific equation, here is a general example: Solve $$\sqrt{x+3} = x - 1$$ 5. Square both sides to eliminate the square root: $$ (\sqrt{x+3})^2 = (x - 1)^2 $$ $$ x + 3 = (x - 1)^2 $$ 6. Expand the right side: $$ x + 3 = x^2 - 2x + 1 $$ 7. Rearrange to form a quadratic equation: $$ 0 = x^2 - 3x - 2 $$ 8. Factor the quadratic: $$ 0 = (x - 2)(x - (-1)) $$ So, $$ x = 2 $$ or $$ x = -1 $$ 9. Check each solution in the original equation: For $$ x = 2 $$: $$ \sqrt{2 + 3} = \sqrt{5} $$ and $$ 2 - 1 = 1 $$ Since $$ \sqrt{5} \neq 1 $$, $$ x = 2 $$ is extraneous. For $$ x = -1 $$: $$ \sqrt{-1 + 3} = \sqrt{2} $$ and $$ -1 - 1 = -2 $$ Since $$ \sqrt{2} \neq -2 $$, $$ x = -1 $$ is also extraneous. 10. In this example, both roots are extraneous, meaning the original equation has no valid solution. Final answer: The extraneous roots are $$ x = 2 $$ and $$ x = -1 $$, and there is no valid solution to the original equation.