To solve the equation \( \sqrt{x+3} = 4 - \sqrt{3x+1} \), let's follow these steps:

### Step 1: Isolate one square root term.
Move one square root term to the other side:
\[
\sqrt{x+3} + \sqrt{3x+1} = 4
\]

### Step 2: Square both sides.
Squaring both sides to eliminate the square roots:
\[
\left( \sqrt{x+3} + \sqrt{3x+1} \right)^2 = 4^2
\]
Expanding both sides:
\[
(x+3) + 2\sqrt{(x+3)(3x+1)} + (3x+1) = 16
\]
Simplify:
\[
x + 3 + 3x + 1 + 2\sqrt{(x+3)(3x+1)} = 16
\]
\[
4x + 4 + 2\sqrt{(x+3)(3x+1)} = 16
\]
Subtract 4 from both sides:
\[
4x + 2\sqrt{(x+3)(3x+1)} = 12
\]

### Step 3: Isolate the square root term.
\[
2\sqrt{(x+3)(3x+1)} = 12 - 4x
\]
Divide by 2:
\[
\sqrt{(x+3)(3x+1)} = 6 - 2x
\]

### Step 4: Square both sides again.
Squaring both sides to eliminate the remaining square root:
\[
(x+3)(3x+1) = (6 - 2x)^2
\]
Expand both sides:
\[
(x+3)(3x+1) = (36 - 24x + 4x^2)
\]
Expanding the left-hand side:
\[
3x^2 + x + 9x + 3 = 36 - 24x + 4x^2
\]
Simplify both sides:
\[
3x^2 + 10x + 3 = 36 - 24x + 4x^2
\]

### Step 5: Move all terms to one side.
Bring all terms to one side to form a quadratic equation:
\[
0 = 4x^2 - 3x^2 - 24x - 10x + 36 - 3
\]
Simplify:
\[
0 = x^2 - 34x + 33
\]

### Step 6: Solve the quadratic equation.
The quadratic equation is:
\[
x^2 - 34x + 33 = 0
\]
Solve using the quadratic formula:
\[
x = \frac{-(-34) \pm \sqrt{(-34)^2 - 4(1)(33)}}{2(1)}
\]
\[
x = \frac{34 \pm \sqrt{1156 - 132}}{2}
\]
\[
x = \frac{34 \pm \sqrt{1024}}{2}
\]
\[
x = \frac{34 \pm 32}{2}
\]
So, the two possible solutions are:
\[
x = \frac{34 + 32}{2} = 33 \quad \text{or} \quad x = \frac{34 - 32}{2} = 1
\]

### Step 7: Check for extraneous solutions.
We need to check both solutions in the original equation:

- For \( x = 33 \):
  \[
  \sqrt{33 + 3} = 4 - \sqrt{3(33) + 1}
  \]
  \[
  \sqrt{36} = 4 - \sqrt{100}
  \]
  \[
  6 = 4 - 10 \quad \text{(false)}
  \]
  So, \( x = 33 \) is not a valid solution.

- For \( x = 1 \):
  \[
  \sqrt{1 + 3} = 4 - \sqrt{3(1) + 1}
  \]
  \[
  \sqrt{4} = 4 - \sqrt{4}
  \]
  \[
  2 = 4 - 2 \quad \text{(true)}
  \]
  So, \( x = 1 \) is the valid solution.

### Final Answer:
The solution to the equation is \( x = 1 \).

Do you want further details or have any questions?

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Here are 5 related questions:
1. How do you handle equations with multiple square roots?
2. Can all quadratic equations have real solutions?
3. What are extraneous solutions, and how do they arise?
4. How does squaring both sides of an equation affect the solution?
5. How do we verify the solutions of equations involving radicals?

**Tip:** Always check your solutions when solving radical equations, as squaring can introduce extraneous solutions.

Learn how to solve the radical equation sqrt(x+3) = 4 - sqrt(3x+1) with a detailed step-by-step solution. This problem involves isolating square root terms, squaring both sides, and solving a quadratic equation. Suitable for students in Grades 9-11.

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