1. **State the problem:** Solve the equation $$\sqrt{4x - 3} = 2x - 1$$ for $x$. 2. **Recall the rule:** To solve equations involving square roots, we isolate the square root term and then square both sides to eliminate the root. 3. **Isolate and square both sides:** $$\sqrt{4x - 3} = 2x - 1$$ Square both sides: $$\left(\sqrt{4x - 3}\right)^2 = (2x - 1)^2$$ which simplifies to $$4x - 3 = (2x - 1)^2$$ 4. **Expand the right side:** $$(2x - 1)^2 = (2x)^2 - 2 \times 2x \times 1 + 1^2 = 4x^2 - 4x + 1$$ 5. **Rewrite the equation:** $$4x - 3 = 4x^2 - 4x + 1$$ 6. **Bring all terms to one side:** $$0 = 4x^2 - 4x + 1 - 4x + 3$$ Simplify: $$0 = 4x^2 - 8x + 4$$ 7. **Divide entire equation by 4 to simplify:** $$0 = \cancel{4}x^2 - \cancel{4} \times 2x + \cancel{4}$$ becomes $$0 = x^2 - 2x + 1$$ 8. **Factor the quadratic:** $$x^2 - 2x + 1 = (x - 1)^2$$ 9. **Solve for $x$:** $$(x - 1)^2 = 0 \implies x - 1 = 0 \implies x = 1$$ 10. **Check for extraneous solutions:** Substitute $x=1$ back into the original equation: $$\sqrt{4(1) - 3} = \sqrt{4 - 3} = \sqrt{1} = 1$$ Right side: $$2(1) - 1 = 2 - 1 = 1$$ Both sides equal 1, so $x=1$ is a valid solution. **Final answer:** $$x = 1$$