To solve the equation $\sqrt{3} + \sqrt{8} = 1 + \sqrt{x}$, let’s go step by step:

---

Step 1: Simplify the radicals

We already know: $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$

So the equation becomes: $\sqrt{3} + 2\sqrt{2} = 1 + \sqrt{x}$

---

Step 2: Isolate $\sqrt{x}$

Rearrange the equation to isolate $\sqrt{x}$ on one side: $\sqrt{x} = \sqrt{3} + 2\sqrt{2} - 1$

---

Step 3: Square both sides

To eliminate the square root, square both sides of the equation: $x = \left( \sqrt{3} + 2\sqrt{2} - 1 \right)^2$

---

Step 4: Expand the squared term

Using the formula $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$, we expand: $\left( \sqrt{3} + 2\sqrt{2} - 1 \right)^2 = (\sqrt{3})^2 + (2\sqrt{2})^2 + (-1)^2 + 2(\sqrt{3})(2\sqrt{2}) - 2(\sqrt{3})(1) - 2(2\sqrt{2})(1)$

Simplify each term:

• $(\sqrt{3})^2 = 3$
• $(2\sqrt{2})^2 = 4 \cdot 2 = 8$
• $(-1)^2 = 1$
• $2(\sqrt{3})(2\sqrt{2}) = 4\sqrt{6}$
• $-2(\sqrt{3})(1) = -2\sqrt{3}$
• $-2(2\sqrt{2})(1) = -4\sqrt{2}$

Add these terms together: $x = 3 + 8 + 1 + 4\sqrt{6} - 2\sqrt{3} - 4\sqrt{2}$

---

Step 5: Final Answer

Combine the terms: $x = 12 + 4\sqrt{6} - 2\sqrt{3} - 4\sqrt{2}$

So the solution is: $x = 12 + 4\sqrt{6} - 2\sqrt{3} - 4\sqrt{2}$

---

Would you like a detailed breakdown of any particular step or help with a related concept?

---

5 Related Questions:

1. How do you simplify expressions with square roots and radicals?
2. What is the process for expanding binomials with radicals?
3. How do you rationalize square root expressions?
4. How do you check if your answer for $x$ satisfies the original equation?
5. Why is squaring both sides a valid method in equations with square roots?

Tip: Always double-check by substituting your solution back into the original equation to verify accuracy!