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What is the solution to [tex]3 x^2 + 5 x \ \textgreater \ -2[/tex]?

A. [tex]x \ \textless \ -1[/tex] or [tex]x \ \textgreater \ -\frac{2}{3}[/tex]
B. [tex]x \geq -\frac{2}{3}[/tex] or [tex]x \ \textless \ 1[/tex]
C. [tex]-\frac{2}{3} \leq x \leq 1[/tex]
D. [tex]-\frac{2}{3} \ \textgreater \ x \ \textgreater \ -1[/tex]


Sagot :

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]