Question

solve the radical equation. $x + 4 = \sqrt{x + 10}$ what is the extraneous solution to the radical equation? options: both $-1$ and $-6$ are true solutions. neither $-1$ nor $-6$ is a true solution to the equation. the solution $-6$ is an extraneous solution.

Explanation:

Step1: Square both sides to eliminate radical

$$(x+4)^2 = (\sqrt{x+10})^2$$
$$x^2 + 8x + 16 = x + 10$$

Step2: Rearrange to quadratic equation

$$x^2 + 8x + 16 - x - 10 = 0$$
$$x^2 + 7x + 6 = 0$$

Step3: Factor the quadratic

$$(x+1)(x+6) = 0$$
$$x = -1 \text{ or } x = -6$$

Step4: Verify solutions in original equation

For $x=-1$:
$$-1 + 4 = \sqrt{-1 + 10}$$
$$3 = \sqrt{9}$$
$$3=3 \text{ (valid)}$$
For $x=-6$:
$$-6 + 4 = \sqrt{-6 + 10}$$
$$-2 = \sqrt{4}$$
$$-2=2 \text{ (invalid)}$$

Answer:

The solution $-6$ is an extraneous solution.