To solve \(|2x + 3| > 3|x + 2|\), consider the non-modular inequality:
\((2x + 3)^2 > 3^2(x + 2)^2\)
Expanding both sides gives:
\((2x + 3)^2 = 4x^2 + 12x + 9\)
\(3^2(x + 2)^2 = 9(x^2 + 4x + 4) = 9x^2 + 36x + 36\)
Set up the inequality:
\(4x^2 + 12x + 9 > 9x^2 + 36x + 36\)
Simplify to:
\(4x^2 + 12x + 9 - 9x^2 - 36x - 36 > 0\)
\(-5x^2 - 24x - 27 > 0\)
Multiply through by -1 (reversing the inequality):
\(5x^2 + 24x + 27 < 0\)
Factor the quadratic:
\((5x + 9)(x + 3) < 0\)
Find critical points by setting each factor to zero:
\(5x + 9 = 0 \Rightarrow x = -\frac{9}{5}\)
\(x + 3 = 0 \Rightarrow x = -3\)
Test intervals around the critical points to determine where the inequality holds:
For \(-3 < x < -\frac{9}{5}\), the inequality holds.
For \(x > -3\) and \(x < -\frac{9}{5}\), the inequality holds.
Thus, the solution is:
\(-3 < x < \frac{9}{5} \text{ or } x > -3 \text{ and } x < \frac{9}{5}\)
