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Find a solution and extraneous solution if any

The equation provided is:

\[
x + 2\sqrt{x} - 15 = 0
\]

Let's solve this equation step by step.

### Step 1: Isolate the square root term
\[
2\sqrt{x} = 15 - x
\]

### Step 2: Square both sides to eliminate the square root
\[
(2\sqrt{x})^2 = (15 - x)^2
\]

\[
4x = 225 - 30x + x^2
\]

### Step 3: Rearrange the equation to form a quadratic equation
\[
x^2 - 34x + 225 = 0
\]

### Step 4: Solve the quadratic equation using the quadratic formula
The quadratic formula is given by:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the quadratic equation \(x^2 - 34x + 225 = 0\), the coefficients are:
- \(a = 1\)
- \(b = -34\)
- \(c = 225\)

Substitute these values into the quadratic formula:

\[
x = \frac{34 \pm \sqrt{(-34)^2 - 4(1)(225)}}{2(1)}
\]

\[
x = \frac{34 \pm \sqrt{1156 - 900}}{2}
\]

\[
x = \frac{34 \pm \sqrt{256}}{2}
\]

\[
x = \frac{34 \pm 16}{2}
\]

So, the possible solutions are:

\[
x = \frac{34 + 16}{2} = \frac{50}{2} = 25
\]
\[
x = \frac{34 - 16}{2} = \frac{18}{2} = 9
\]

### Step 5: Check for extraneous solutions
We need to substitute both values back into the original equation to ensure they satisfy it.

1. **For \(x = 25\):**

\[
25 + 2\sqrt{25} - 15 = 25 + 10 - 15 = 20 \quad (\text{not } 0)
\]

This is an extraneous solution.

2. **For \(x = 9\):**

\[
9 + 2\sqrt{9} - 15 = 9 + 6 - 15 = 0
\]

This solution satisfies the original equation.

### Conclusion
The valid solution is \(x = 9\). The extraneous solution is \(x = 25\).

Would you like more details or have any questions?

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Here are 5 related questions:
1. How can extraneous solutions arise in equations involving square roots?
2. What is the significance of checking for extraneous solutions?
3. How can you solve equations involving higher powers and roots?
4. What are the conditions that make a quadratic equation have real roots?
5. How can the quadratic formula be derived?

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

Find the solution and extraneous solution if any for the equation x + 2√x - 15 = 0.

This problem demonstrates how to solve the radical equation x + 2√x - 15 = 0, including the use of the quadratic equation formula and identifying extraneous solutions. Suitable for students in Grades 9-11.

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