Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

square root of the quantity x plus 6 end quantity minus 4 equals x

We are asked to solve the equation:

\[
\sqrt{x + 6} - 4 = x
\]

Let's break down the steps carefully to solve for \(x\).

### Step 1: Isolate the square root term
First, add 4 to both sides of the equation to isolate the square root term:

\[
\sqrt{x + 6} = x + 4
\]

### Step 2: Square both sides
To eliminate the square root, square both sides of the equation:

\[
(\sqrt{x + 6})^2 = (x + 4)^2
\]

Simplifying both sides:

\[
x + 6 = (x + 4)^2
\]

### Step 3: Expand the right-hand side
Now expand \( (x + 4)^2 \):

\[
x + 6 = (x + 4)(x + 4) = x^2 + 8x + 16
\]

### Step 4: Move all terms to one side
To form a quadratic equation, subtract \(x + 6\) from both sides:

\[
0 = x^2 + 8x + 16 - x - 6
\]

Simplify the terms:

\[
0 = x^2 + 7x + 10
\]

### Step 5: Solve the quadratic equation
Now we solve the quadratic equation \( x^2 + 7x + 10 = 0 \). We can factor this equation:

\[
x^2 + 7x + 10 = (x + 5)(x + 2) = 0
\]

Setting each factor equal to zero:

\[
x + 5 = 0 \quad \text{or} \quad x + 2 = 0
\]

So, the possible solutions are:

\[
x = -5 \quad \text{or} \quad x = -2
\]

### Step 6: Check for extraneous solutions
We need to check both solutions by substituting them back into the original equation to see if they produce valid results.

**Check \(x = -5\):**

Substitute \(x = -5\) into the original equation:

\[
\sqrt{-5 + 6} - 4 = -5
\]
\[
\sqrt{1} - 4 = -5
\]
\[
1 - 4 = -3
\]

This does **not** equal \(-5\), so \(x = -5\) is an **extraneous solution**.

**Check \(x = -2\):**

Substitute \(x = -2\) into the original equation:

\[
\sqrt{-2 + 6} - 4 = -2
\]
\[
\sqrt{4} - 4 = -2
\]
\[
2 - 4 = -2
\]

This is true, so \(x = -2\) is a valid solution.

### Final Answer:
The only solution is:

\[
x = -2
\]

Would you like further details on any part of the solution?

---

Here are 5 related questions to expand on this concept:

1. How do you handle extraneous solutions in equations involving square roots?
2. What other types of equations produce extraneous solutions, and why?
3. Can you solve \( \sqrt{2x + 5} - 3 = x \) and check for extraneous solutions?
4. How would you approach solving an equation with multiple square roots?
5. What does squaring both sides of an equation do to the solution set?

**Tip**: Always check for extraneous solutions when solving equations involving square roots, as squaring both sides can introduce invalid solutions.

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer. Square root of the quantity x plus 6 end quantity minus 4 equals x.

Step-by-step solution to solve the equation with square roots: sqrt(x + 6) - 4 = x. This problem explores algebraic concepts such as isolating square roots, solving quadratic equations, and checking for extraneous solutions. Ideal for students in Grades 9-11.

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