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}

Sat, 02 May 2026 15:33:37 +0000

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