Solve the inequality:

$|2x + 5| \leq 3|2x - 1|$

To solve $|2x + 5| \leq 3|2x - 1|$, square both sides (since absolute values are non-negative):

$|2x + 5|^2 \leq 9|2x - 1|^2$

$(2x + 5)^2 \leq 9(2x - 1)^2$

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Expanding both sides:

$4x^2 + 20x + 25 \leq 9(4x^2 - 4x + 1)$

$4x^2 + 20x + 25 \leq 36x^2 - 36x + 9$

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Rearranging:

$0 \leq 36x^2 - 4x^2 - 36x - 20x + 9 - 25$

$0 \leq 32x^2 - 56x - 16$

$0 \leq 8(4x^2 - 7x - 2)$

$4x^2 - 7x - 2 \geq 0$

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Finding roots using the quadratic formula:

$x = \frac{7 \pm \sqrt{49 + 32}}{8}$

$x = \frac{7 \pm \sqrt{81}}{8}$

$x = \frac{7 \pm 9}{8}$

Therefore $x = 2$ or $x = -\frac{1}{4}$

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Factoring the quadratic:

$4x^2 - 7x - 2 = 4(x - 2)(x + \frac{1}{4})$

For $(x - 2)(x + \frac{1}{4}) \geq 0$, both factors must have the same sign.

When $x \geq 2$: both factors are positive ✓

When $x \leq -\frac{1}{4}$: both factors are negative ✓

When $-\frac{1}{4} < x < 2$: factors have opposite signs ✗

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Therefore, the solution is:

$x \leq -\frac{1}{4}$ or $x \geq 2$

The correct answer is Option 4.