Solve the inequality 5a^2 + 4a - 12 > 0.

Step-by-step solution

1. 1

   Find the roots of the associated equation. Solve $5a^2+4a-12=0$ with the quadratic formula, $a$-coefficients $5$, $4$, $-12$:

   $\Delta=4^2-4(5)(-12)=16+240=256$

2. 2

   Take advantage of the perfect square. $\sqrt{256}=16$, so the roots are rational:

   $a=\frac{-4\pm 16}{10}\quad\Longrightarrow\quad a=\frac{12}{10}=\frac{6}{5}\ \text{or}\ a=\frac{-20}{10}=-2$

   Equivalently $5a^2+4a-12=(5a-6)(a+2)$.

3. 3

   Use the direction of opening. The leading coefficient $5$ is positive, so the parabola opens upward and the expression is positive outside the roots and negative between them.

4. 4

   Order the roots and write the answer. Since $-2<\tfrac65$:

   $a<-2\quad\text{or}\quad a>\frac{6}{5},\qquad\text{i.e. }(-\infty,-2)\cup\left(\tfrac65,\infty\right)$

5. 5

   Check one value per interval. At $a=-3$: $45-12-12=21>0$ $\checkmark$. At $a=0$: $-12>0$ false $\checkmark$. At $a=2$: $20+8-12=16>0$ $\checkmark$. At $a=\tfrac65$: $5\cdot\tfrac{36}{25}+\tfrac{24}{5}-12=\tfrac{36}{5}+\tfrac{24}{5}-12=0$, correctly excluded by the strict inequality.