To solve the inequality \(|3x - a| > 2|x + 2a|\), we first consider the non-modular inequality:
\((3x - a)^2 > 2^2(x + 2a)^2\).
This simplifies to:
\((3x - a)^2 > 4(x + 2a)^2\).
Expanding both sides, we get:
\(9x^2 - 6ax + a^2 > 4(x^2 + 4ax + 4a^2)\).
Simplifying further:
\(9x^2 - 6ax + a^2 > 4x^2 + 16ax + 16a^2\).
Rearranging terms gives:
\(5x^2 - 22ax - 15a^2 > 0\).
We solve the quadratic equation \(5x^2 - 22ax - 15a^2 = 0\) to find critical values:
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b = -22a\), \(c = -15a^2\), we find:
\(x = \frac{22a \pm \sqrt{(-22a)^2 - 4 \cdot 5 \cdot (-15a^2)}}{2 \cdot 5}\).
\(x = \frac{22a \pm \sqrt{484a^2 + 300a^2}}{10}\).
\(x = \frac{22a \pm \sqrt{784a^2}}{10}\).
\(x = \frac{22a \pm 28a}{10}\).
This gives critical values \(x = 5a\) and \(x = -\frac{3}{5}a\).
The solution to the inequality is:
\(x > 5a\) or \(x < -\frac{3}{5}a\).
