Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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]
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]
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.