Solve for all values of x : \[ \sqrt{3 x}+2=\sqrt{6 x-1} \]

To solve the equation \( \sqrt{3x} + 2 = \sqrt{6x - 1} \), we start by isolating one of the square root terms: 1. Move 2 to the other side: \[ \sqrt{3x} = \sqrt{6x - 1} - 2 \] 2. Square both sides to eliminate the square roots: \[ (\sqrt{3x})^2 = (\sqrt{6x - 1} - 2)^2 \] This simplifies to: \[ 3x = (6x - 1) - 4\sqrt{6x - 1} + 4 \] Which further simplifies to: \[ 3x = 6x + 3 - 4\sqrt{6x - 1} \] 3. Rearrange the equation to isolate the square root: \[ 4\sqrt{6x - 1} = 6x + 3 - 3x \] Simplifying gives: \[ 4\sqrt{6x - 1} = 3x + 3 \] 4. Divide both sides by 4: \[ \sqrt{6x - 1} = \frac{3x + 3}{4} \] 5. Square both sides again: \[ 6x - 1 = \left( \frac{3x + 3}{4} \right)^2 \] This expands to: \[ 6x - 1 = \frac{(3x + 3)^2}{16} \] Simplifying the right side: \[ 6x - 1 = \frac{9x^2 + 18x + 9}{16} \] 6. Multiply everything by 16 to eliminate the fraction: \[ 16(6x - 1) = 9x^2 + 18x + 9 \] Simplifying gives: \[ 96x - 16 = 9x^2 + 18x + 9 \] 7. Rearranging the equation: \[ 0 = 9x^2 - 78x + 25 \] We can now use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 9, b = -78, c = 25 \): Calculate the discriminant: \[ b^2 - 4ac = (-78)^2 - 4 \cdot 9 \cdot 25 = 6084 - 900 = 5184 \] Now applying the quadratic formula: \[ x = \frac{78 \pm \sqrt{5184}}{18} \] Since \( \sqrt{5184} = 72 \): \[ x = \frac{78 \pm 72}{18} \] 8. Solving for the two possible values of \( x \): \[ x_1 = \frac{150}{18} = \frac{25}{3} \] \[ x_2 = \frac{6}{18} = \frac{1}{3} \] Finally, we check for extraneous solutions by substituting back into the original equation: For \( x = \frac{25}{3} \): \[ \sqrt{3 \cdot \frac{25}{3}} + 2 = \sqrt{6 \cdot \frac{25}{3} - 1} \implies \sqrt{25} + 2 = \sqrt{50 - 1} \implies 5 + 2 = \sqrt{49} \implies 7 = 7 \quad \text{(valid)} \] For \( x = \frac{1}{3} \): \[ \sqrt{3 \cdot \frac{1}{3}} + 2 = \sqrt{6 \cdot \frac{1}{3} - 1} \implies \sqrt{1} + 2 = \sqrt{2 - 1} \implies 1 + 2 = \sqrt{1} \implies 3 = 1 \quad \text{(not valid)} \] Thus, the only solution is: \[ \boxed{\frac{25}{3}} \]