Question: 6$x^2$ +x-2$\le$ 0
1. **State the problem:** Solve the inequality $$6x^2 + x - 2 \le 0$$.
2. **Identify the type of inequality:** This is a quadratic inequality of the form $$ax^2 + bx + c \le 0$$ where $$a=6$$, $$b=1$$, and $$c=-2$$.
3. **Find the roots of the quadratic equation:** Solve $$6x^2 + x - 2 = 0$$ using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Substitute values:
$$x = \frac{-1 \pm \sqrt{1^2 - 4 \times 6 \times (-2)}}{2 \times 6} = \frac{-1 \pm \sqrt{1 + 48}}{12} = \frac{-1 \pm \sqrt{49}}{12}$$
$$x = \frac{-1 \pm 7}{12}$$
4. **Calculate the two roots:**
$$x_1 = \frac{-1 - 7}{12} = \frac{-8}{12} = -\frac{2}{3}$$
$$x_2 = \frac{-1 + 7}{12} = \frac{6}{12} = \frac{1}{2}$$
5. **Determine the intervals to test:** The roots divide the number line into three intervals:
- $$(-\infty, -\frac{2}{3})$$
- $$[-\frac{2}{3}, \frac{1}{2}]$$
- $$(\frac{1}{2}, \infty)$$
6. **Test the sign of the quadratic in each interval:** Since $$a=6 > 0$$, the parabola opens upwards.
- For $$x$$ less than $$-\frac{2}{3}$$, the quadratic is positive.
- Between $$-\frac{2}{3}$$ and $$\frac{1}{2}$$, the quadratic is less than or equal to zero.
- For $$x$$ greater than $$\frac{1}{2}$$, the quadratic is positive.
7. **Write the solution:**
$$\boxed{-\frac{2}{3} \le x \le \frac{1}{2}}$$
This is the set of $$x$$ values that satisfy the inequality.