1. **State the problem:** Solve the equation $$6 \sqrt{4x} - 3 = 2x - 1$$ for $x$. 2. **Isolate the square root term:** Add 3 to both sides: $$6 \sqrt{4x} = 2x - 1 + 3$$ $$6 \sqrt{4x} = 2x + 2$$ 3. **Divide both sides by 6 to simplify:** $$\cancel{6} \sqrt{4x} = \frac{2x + 2}{\cancel{6}}$$ $$\sqrt{4x} = \frac{2x + 2}{6} = \frac{x + 1}{3}$$ 4. **Square both sides to eliminate the square root:** $$\left(\sqrt{4x}\right)^2 = \left(\frac{x + 1}{3}\right)^2$$ $$4x = \frac{(x + 1)^2}{9}$$ 5. **Multiply both sides by 9 to clear the denominator:** $$9 \times 4x = 9 \times \frac{(x + 1)^2}{9}$$ $$36x = (x + 1)^2$$ 6. **Expand the right side:** $$(x + 1)^2 = x^2 + 2x + 1$$ So, $$36x = x^2 + 2x + 1$$ 7. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 + 2x + 1 - 36x$$ $$0 = x^2 - 34x + 1$$ 8. **Use the quadratic formula to solve for $x$:** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-34$, and $c=1$. Calculate the discriminant: $$b^2 - 4ac = (-34)^2 - 4 \times 1 \times 1 = 1156 - 4 = 1152$$ Calculate the square root: $$\sqrt{1152} = \sqrt{256 \times 4.5} = 16 \sqrt{4.5} = 16 \times \sqrt{\frac{9}{2}} = 16 \times \frac{3}{\sqrt{2}} = \frac{48}{\sqrt{2}} = 24 \sqrt{2}$$ So, $$x = \frac{34 \pm 24 \sqrt{2}}{2} = 17 \pm 12 \sqrt{2}$$ 9. **Check for extraneous solutions:** Substitute $x = 17 + 12 \sqrt{2}$ into the original equation: - Calculate $\sqrt{4x} = \sqrt{4(17 + 12 \sqrt{2})} = \sqrt{68 + 48 \sqrt{2}}$ which is positive. - Both sides will be equal, so this solution is valid. Substitute $x = 17 - 12 \sqrt{2}$: - Calculate $\sqrt{4x} = \sqrt{4(17 - 12 \sqrt{2})} = \sqrt{68 - 48 \sqrt{2}}$ which is positive. - Check if both sides equal; this also holds. 10. **Final answer:** $$x = 17 + 12 \sqrt{2} \quad \text{or} \quad x = 17 - 12 \sqrt{2}$$