We are solving the inequality:

$\sqrt{4x - 3} \geq 2x - 3$

Step 1: Analyze the Domain

For the square root to be defined, the expression inside the square root must be non-negative: $4x - 3 \geq 0 \implies x \geq \frac{3}{4}.$

Step 2: Solve the Inequality

Square both sides to eliminate the square root (be cautious with squaring since it might introduce extraneous solutions):

$\sqrt{4x - 3} \geq 2x - 3 \implies ( \sqrt{4x - 3})^2 \geq (2x - 3)^2$

$4x - 3 \geq 4x^2 - 12x + 9$

Simplify: $0 \geq 4x^2 - 16x + 12$

Rearrange: $4x^2 - 16x + 12 \leq 0$

Step 3: Solve the Quadratic Inequality

Factorize the quadratic equation: $4(x^2 - 4x + 3) \leq 0$

$4(x - 3)(x - 1) \leq 0$

The critical points are $x = 1$ and $x = 3$. To determine the intervals where the inequality holds, test the signs of $4(x - 3)(x - 1)$ in each interval: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

• For $x \in (1, 3)$, $4(x - 3)(x - 1) < 0$.
• For $x \in (-\infty, 1)$ or $x \in (3, \infty)$, $4(x - 3)(x - 1) > 0$.

Thus, $x \in [1, 3]$.

Step 4: Verify Against the Domain

From Step 1, $x \geq \frac{3}{4}$. Combine this with $x \in [1, 3]$: $x \in [1, 3].$

Final Answer:

The solution is: $x \in [1, 3].$

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Would you like further details or have questions? Here are 5 related problems to practice:

1. Solve $\sqrt{3x + 2} \leq x + 1$.
2. Solve $\sqrt{5x - 1} = 2x - 3$.
3. Solve $\sqrt{2x + 4} > x - 1$.
4. Solve $x^2 - 5x + 6 \leq 0$ and interpret its critical points.
5. Solve the quadratic inequality $x^2 - 7x + 12 > 0$.

Tip:

Always check for extraneous solutions when squaring both sides of an equation or inequality!