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Solve the radical equation. Check all proposed solutions.

StartRoot x plus 13 EndRoot equals x minus 7x+13=x−7

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Part 1

Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

A.

The solution set is StartSet nothing EndSet .enter your response here.

​(Use a comma to separate answers as needed. Simplify your​ answer.)

B.

The solution set is the empty set.

To solve the given radical equation:

\[
\sqrt{x + 13} = x - 7
\]

we will follow these steps systematically:

### Step 1: Square Both Sides

Square both sides of the equation to eliminate the square root:

\[
(\sqrt{x + 13})^2 = (x - 7)^2
\]

Simplifying, we get:

\[
x + 13 = (x - 7)^2
\]

### Step 2: Expand the Right Side

Expand the right-hand side:

\[
x + 13 = x^2 - 14x + 49
\]

### Step 3: Rearrange into a Quadratic Equation

Bring all terms to one side to form a standard quadratic equation:

\[
x^2 - 14x - x + 49 - 13 = 0
\]

Simplify the terms:

\[
x^2 - 15x + 36 = 0
\]

### Step 4: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -15 \), and \( c = 36 \).

\[
x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(36)}}{2(1)}
\]

Simplify:

\[
x = \frac{15 \pm \sqrt{225 - 144}}{2}
\]

\[
x = \frac{15 \pm \sqrt{81}}{2}
\]

\[
x = \frac{15 \pm 9}{2}
\]

This gives two solutions:

\[
x = \frac{15 + 9}{2} = 12 \quad \text{and} \quad x = \frac{15 - 9}{2} = 3
\]

### Step 5: Check Both Solutions in the Original Equation

1. For \( x = 12 \):

   \[
   \sqrt{12 + 13} = \sqrt{25} = 5
   \]

   And \( x - 7 = 12 - 7 = 5 \). Since both sides match, \( x = 12 \) is a solution.

2. For \( x = 3 \):

   \[
   \sqrt{3 + 13} = \sqrt{16} = 4
   \]

   And \( x - 7 = 3 - 7 = -4 \). Since both sides do not match, \( x = 3 \) is not a valid solution.

### Final Answer

The solution set is:

\[
\text{The solution set is } \{12\}.
\]

Would you like further details or have any questions? Here are some related questions you might explore:

1. How do extraneous solutions arise in radical equations?
2. What are the steps for isolating radicals in equations?
3. Can a radical equation have more than one valid solution?
4. How do you verify if a solution is extraneous?
5. What happens if the radical is isolated on the other side of the equation?

**Tip:** Always check each solution by substituting it back into the original equation, especially when squaring both sides, as it can introduce extraneous solutions that are not valid.

Learn how to solve the radical equation sqrt(x + 13) = x - 7 with a detailed step-by-step solution. This problem involves radical equations and quadratic equations, requiring application of the quadratic formula. Suitable for students in Grades 10-12.

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