Find detailed and accurate answers to your questions on IDNLearn.com. Ask any question and get a detailed, reliable answer from our community of experts.
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]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.