To solve the equation \(\sqrt{x + 19} + 1 = \sqrt{2x + 4}\), follow these steps:

1. **Isolate one of the square roots:**
   \[
   \sqrt{x + 19} = \sqrt{2x + 4} - 1
   \]

2. **Square both sides to eliminate the square roots:**
   \[
   (\sqrt{x + 19})^2 = (\sqrt{2x + 4} - 1)^2
   \]
   \[
   x + 19 = (2x + 4) - 2\sqrt{2x + 4} + 1
   \]
   \[
   x + 19 = 2x + 5 - 2\sqrt{2x + 4}
   \]

3. **Isolate the term with the square root:**
   \[
   x + 19 - 2x - 5 = -2\sqrt{2x + 4}
   \]
   \[
   -x + 14 = -2\sqrt{2x + 4}
   \]
   \[
   x - 14 = 2\sqrt{2x + 4}
   \]

4. **Square both sides again to eliminate the remaining square root:**
   \[
   (x - 14)^2 = (2\sqrt{2x + 4})^2
   \]
   \[
   x^2 - 28x + 196 = 4(2x + 4)
   \]
   \[
   x^2 - 28x + 196 = 8x + 16
   \]

5. **Move all terms to one side to form a quadratic equation:**
   \[
   x^2 - 28x + 196 - 8x - 16 = 0
   \]
   \[
   x^2 - 36x + 180 = 0
   \]

6. **Solve the quadratic equation:**
   The quadratic equation \(x^2 - 36x + 180 = 0\) can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -36\), and \(c = 180\):
   \[
   x = \frac{-(-36) \pm \sqrt{(-36)^2 - 4 \cdot 1 \cdot 180}}{2 \cdot 1}
   \]
   \[
   x = \frac{36 \pm \sqrt{1296 - 720}}{2}
   \]
   \[
   x = \frac{36 \pm \sqrt{576}}{2}
   \]
   \[
   x = \frac{36 \pm 24}{2}
   \]

   Therefore, the solutions are:
   \[
   x = \frac{36 + 24}{2} = 30 \quad \text{and} \quad x = \frac{36 - 24}{2} = 6
   \]

7. **Verify the solutions:**
   - For \(x = 30\):
     \[
     \sqrt{30 + 19} + 1 = \sqrt{2 \cdot 30 + 4}
     \]
     \[
     \sqrt{49} + 1 = \sqrt{64}
     \]
     \[
     7 + 1 = 8
     \]
     \[
     8 = 8 \quad \text{(True)}
     \]

   - For \(x = 6\):
     \[
     \sqrt{6 + 19} + 1 = \sqrt{2 \cdot 6 + 4}
     \]
     \[
     \sqrt{25} + 1 = \sqrt{16}
     \]
     \[
     5 + 1 = 4
     \]
     \[
     6 \neq 4 \quad \text{(False)}
     \]

Therefore, the only solution is:
\[
x = 30
\]

Would you like more details or have any questions?

Here are 5 questions you might want to ask:
1. How do you simplify expressions involving square roots?
2. Can you explain the steps of solving a quadratic equation?
3. What are common mistakes to avoid when solving equations with square roots?
4. How can I verify solutions to an equation?
5. What are the properties of square roots that are useful in solving equations?

**Tip:** Always verify potential solutions to ensure they satisfy the original equation, especially when squaring both sides can introduce extraneous solutions.

Learn how to solve the equation √𝑥+19+1=√2𝑥+4 with a detailed step-by-step solution. This problem covers equations involving square roots and quadratic equations, suitable for advanced high school students (Grades 10-12).

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