To solve the inequality [tex]\(3 x^2 + 5 x > -2\)[/tex], we can follow these steps:
1. Rewrite the inequality:
[tex]\[
3 x^2 + 5 x + 2 > 0
\][/tex]
This inequality is now a quadratic inequality.
2. Find the roots of the quadratic equation [tex]\(3 x^2 + 5 x + 2 = 0\)[/tex]:
To do this, we can use the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\(a = 3\)[/tex], [tex]\(b = 5\)[/tex], and [tex]\(c = 2\)[/tex].
First, calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[
\Delta = b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1
\][/tex]
The discriminant is 1, which is positive, indicating two real and distinct roots.
Hence, the roots are:
[tex]\[
x = \frac{-5 \pm \sqrt{1}}{2 \cdot 3} = \frac{-5 \pm 1}{6}
\][/tex]
Thus, the roots are:
[tex]\[
x = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3}
\][/tex]
and
[tex]\[
x = \frac{-5 - 1}{6} = \frac{-6}{6} = -1
\][/tex]
3. Determine the sign of [tex]\(3 x^2 + 5 x + 2\)[/tex] in the intervals determined by the roots:
The roots [tex]\(-1\)[/tex] and [tex]\(-\frac{2}{3}\)[/tex] divide the number line into three intervals: [tex]\((-\infty, -1)\)[/tex], [tex]\((-1, -\frac{2}{3})\)[/tex], and [tex]\((-\frac{2}{3}, \infty)\)[/tex].
- For [tex]\(x \in (-\infty, -1)\)[/tex]: Choose [tex]\(x = -2\)[/tex]:
[tex]\[
3(-2)^2 + 5(-2) + 2 = 12 - 10 + 2 = 4 > 0
\][/tex]
- For [tex]\(x \in (-1, -\frac{2}{3})\)[/tex]: Choose [tex]\(x = -\frac{3}{4}\)[/tex]:
[tex]\[
3\left(-\frac{3}{4}\right)^2 + 5\left(-\frac{3}{4}\right) + 2 = 3 \cdot \frac{9}{16} - \frac{15}{4} + 2 = \frac{27}{16} - \frac{60}{16} + \frac{32}{16} = -\frac{1}{16} < 0
\][/tex]
- For [tex]\(x \in (-\frac{2}{3}, \infty)\)[/tex]: Choose [tex]\(x = 0\)[/tex]:
[tex]\[
3(0)^2 + 5(0) + 2 = 2 > 0
\][/tex]
4. Combine the intervals where the quadratic expression is positive:
From the above analysis:
- [tex]\(3 x^2 + 5 x + 2 > 0\)[/tex] in the intervals [tex]\((-\infty, -1)\)[/tex] and [tex]\((-\frac{2}{3}, \infty)\)[/tex].
5. Write down the solution in interval notation:
[tex]\[
x \in (-\infty, -1) \cup (-\frac{2}{3}, \infty)
\][/tex]
6. Identify the corresponding option:
The correct solution matches the first option: [tex]\(x <-1\)[/tex] or [tex]\(x > -\frac{2}{3}\)[/tex].
Thus, the solution to the inequality [tex]\(3 x^2 + 5 x > -2\)[/tex] is:
[tex]\[ x < -1 \text{ or } x > -\frac{2}{3} \][/tex]
