We are given the inequality:

$\frac{\sqrt{5x + 4} - 3}{2x} \geq 1$

Step 1: Isolate the square root

Multiply both sides by $2x$ (assuming $x \neq 0$):

$\sqrt{5x + 4} - 3 \geq 2x$

Step 2: Move terms around

Add 3 to both sides:

$\sqrt{5x + 4} \geq 2x + 3$

Step 3: Square both sides

Square both sides to eliminate the square root:

$5x + 4 \geq (2x + 3)^2$

Now expand the right side:

$5x + 4 \geq 4x^2 + 12x + 9$

Step 4: Rearrange the inequality

Move all terms to one side to set the inequality to 0:

$0 \geq 4x^2 + 12x + 9 - 5x - 4$

Simplify:

$0 \geq 4x^2 + 7x + 5$

or

$4x^2 + 7x + 5 \leq 0$

Step 5: Solve the quadratic inequality

To solve $4x^2 + 7x + 5 \leq 0$, we first find the roots of the corresponding quadratic equation $4x^2 + 7x + 5 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4$, $b = 7$, and $c = 5$. Plugging in these values:

$x = \frac{-7 \pm \sqrt{7^2 - 4(4)(5)}}{2(4)}$ $x = \frac{-7 \pm \sqrt{49 - 80}}{8}$ $x = \frac{-7 \pm \sqrt{-31}}{8}$

Since the discriminant ($\sqrt{-31}$) is negative, there are no real solutions for this quadratic equation. This means that $4x^2 + 7x + 5$ is always positive, and it does not satisfy $\leq 0$.

Step 6: Consider the domain

The expression $\sqrt{5x + 4}$ requires that $5x + 4 \geq 0$, so:

$x \geq -\frac{4}{5}$

Final Answer:

The inequality has no real solution since the quadratic inequality $4x^2 + 7x + 5 \leq 0$ has no real solutions, and the expression $\sqrt{5x + 4}$ is defined for $x \geq -\frac{4}{5}$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you solve inequalities involving square roots?
2. What happens if a quadratic equation has no real roots?
3. Can inequalities have no solutions? How do you determine that?
4. How do you interpret the domain when dealing with square roots?
5. What are the implications of a negative discriminant in a quadratic equation?

Tip: Always check the domain of expressions with square roots before solving!