Solution: Let $ u = \sqrt{x} $, so $ u \geq 0 $. Substituting, the equation becomes $ u + 3u^2 = 4 $. Rearranging: $ 3u^2 + u - 4 = 0 $. Using the quadratic formula: $ u = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6} $. Valid solution: $ u = 1 $ (since $ u \geq 0 $). Thus, $ \sqrt{x} = 1 \Rightarrow x = 1 $. \boxed{1}

["Solving Quadratic Equations Using Substitution: A Step-by-Step Approach", "When faced with an equation involving square roots, substitution can simplify the problem and transform it into a more familiar quadratic form. Consider the equation:", "$$\n\sqrt{x} + 3x = 4\n$$", "This equation includes a radical, making direct solving cumbersome. However, by letting $ u = \sqrt{x} $, where $ u \geq 0 $ (since square roots are non-negative), we streamline the process.", "---", "### Step 1: Substitute $ u = \sqrt{x} $", "Replacing $ \sqrt{x} $ with $ u $, and noting that $ x = u^2 $, the equation becomes:", "$$\nu + 3u^2 = 4\n$$", "---", "### Step 2: Rearrange into standard quadratic form", "Bring all terms to one side:", "$$\n3u^2 + u - 4 = 0\n$$", "---", "### Step 3: Apply the quadratic formula", "For any quadratic equation $ au^2 + bu + c = 0 $, the solutions are given by:", "$$\nu = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n$$", "Here, $ a = 3 $, $ b = 1 $, $ c = -4 $. Substituting:", "$$\nu = \frac{-1 \pm \sqrt{1^2 - 4(3)(-4)}}{2 \cdot 3} = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm \sqrt{49}}{6}\n$$", "$$\nu = \frac{-1 \pm 7}{6}\n$$", "So the two solutions are:", "$$\nu = \frac{6}{6} = 1 \quad \ ext{and} \quad u = \frac{-8}{6} = -\frac{4}{3}\n$$", "---", "### Step 4: Select the valid solution", "Since $ u = \sqrt{x} \geq 0 $, we discard the negative $ u = -\frac{4}{3} $. Thus, the only valid solution is:", "$$\nu = 1\n$$", "---", "### Step 5: Solve for $ x $", "Recall $ x = u^2 $, so:", "$$\nx = 1^2 = 1\n$$", "---", "### Final Answer", "$$\n\boxed{1}\n$$", "---", "Why This Method Works:\nSubstitution turns nonlinear equations with square roots into solvable quadratics, especially when combined with domain awareness. Recognizing when to restrict variables (like $ u \geq 0 $) ensures only physically or mathematically valid solutions are accepted. This strategy applies broadly in algebra, calculus, and applied mathematics."]