Question

algebra: concepts and connections - plo radical equations consider the radical equation \\(\sqrt{n + 4} = n - 2\\). which statement is true about the solutions \\(n = 5\\) and \\(n = 0\\)? the solution \\(n = 5\\) is an extraneous solution. neither are true solutions to the equation. the solution \\(n = 0\\) is an extraneous solution. both \\(n = 5\\) and \\(n = 0\\) are true solutions.

Explanation:

Step1: Square both sides

$$(\sqrt{n+4})^2=(n-2)^2$$
$$n+4=n^2-4n+4$$

Step2: Rearrange to quadratic form

$$n^2-5n=0$$
$$n(n-5)=0$$

Step3: Solve for n

$$n=0 \text{ or } n=5$$

Step4: Check n=0 in original equation

$$\sqrt{0+4}=0-2$$
$$2=-2 \quad \text{(False, extraneous)}$$

Step5: Check n=5 in original equation

$$\sqrt{5+4}=5-2$$
$$3=3 \quad \text{(True, valid)}$$

Answer:

The solution $n = 0$ is an extraneous solution.