https://file.aiquickdraw.com/mathbot/file/1728573942919bhj1wy1m.jpg

Let's solve the equation step by step:

\[
\sqrt{19 - x} = x + 1
\]

### Step 1: Square both sides of the equation
To eliminate the square root, square both sides:

\[
(\sqrt{19 - x})^2 = (x + 1)^2
\]

This simplifies to:

\[
19 - x = (x + 1)^2
\]

### Step 2: Expand the right side
Now, expand \( (x + 1)^2 \):

\[
19 - x = x^2 + 2x + 1
\]

### Step 3: Rearrange the equation
Move all terms to one side of the equation:

\[
0 = x^2 + 2x + 1 + x - 19
\]

Simplify:

\[
0 = x^2 + 3x - 18
\]

### Step 4: Solve the quadratic equation
We now solve \( x^2 + 3x - 18 = 0 \). This can be factored as:

\[
(x + 6)(x - 3) = 0
\]

So, the solutions are:

\[
x = -6 \quad \text{or} \quad x = 3
\]

### Step 5: Check for extraneous solutions
Since the original equation involves a square root, we need to check both solutions in the original equation to determine if either is extraneous.

- **For \( x = -6 \):**
  Substituting into the original equation:

  \[
  \sqrt{19 - (-6)} = -6 + 1
  \]

  Simplifies to:

  \[
  \sqrt{25} = -5
  \]

  But this is false because \( \sqrt{25} = 5 \), not \(-5\). So, \( x = -6 \) is an extraneous solution.

- **For \( x = 3 \):**
  Substituting into the original equation:

  \[
  \sqrt{19 - 3} = 3 + 1
  \]

  Simplifies to:

  \[
  \sqrt{16} = 4
  \]

  This is true because \( \sqrt{16} = 4 \). So, \( x = 3 \) is a valid solution.

### Final Answer:
The solution to the equation is \( x = 3 \), and the extraneous solution is \( x = -6 \).

Would you like further details or have any questions about this? Here are some additional related questions:

1. How do you solve square root equations in general?
2. Why do extraneous solutions occur in equations involving square roots?
3. What other methods can be used to solve quadratic equations?
4. Can you explain how the factoring method works step-by-step?
5. How can we check for extraneous solutions in other types of equations?

**Tip:** Always check your solutions when dealing with square roots or rational equations to ensure there are no extraneous solutions.

Solve and identify the extraneous solution (if there is one): sqrt(19 - x) = x + 1.

Learn how to solve the square root equation sqrt(19 - x) = x + 1 step-by-step. This problem involves algebraic concepts like quadratic equations and extraneous solutions. Suitable for Grades 9-12.

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