Discover new knowledge and insights with IDNLearn.com's extensive Q&A database. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
To determine the values of [tex]\( x \)[/tex] that satisfy the inequality [tex]\(\frac{x^2 + x - 2}{x^2 + 4} > \frac{1}{2}\)[/tex], we can follow these steps:
1. Rewrite the Inequality:
First, let's rewrite the given inequality:
[tex]\[ \frac{x^2 + x - 2}{x^2 + 4} > \frac{1}{2} \][/tex]
2. Combine the Fractions:
To eliminate the fraction on the right-hand side, we need a common denominator:
[tex]\[ \frac{x^2 + x - 2}{x^2 + 4} - \frac{1}{2} > 0 \][/tex]
Let's combine the fractions:
[tex]\[ \frac{2(x^2 + x - 2) - (x^2 + 4)}{2(x^2 + 4)} > 0 \][/tex]
3. Simplify the Numerator:
Simplify the numerator inside the fraction:
[tex]\[ \frac{2x^2 + 2x - 4 - x^2 - 4}{2(x^2 + 4)} > 0 \][/tex]
Combining like terms, we get:
[tex]\[ \frac{x^2 + 2x - 8}{2(x^2 + 4)} > 0 \][/tex]
4. Analyze the Simplified Inequality:
Now, we have:
[tex]\[ \frac{x^2 + 2x - 8}{2(x^2 + 4)} > 0 \][/tex]
5. Factor the Numerator:
Factorize the numerator [tex]\( x^2 + 2x - 8 \)[/tex]:
[tex]\[ x^2 + 2x - 8 = (x - 2)(x + 4) \][/tex]
So the inequality becomes:
[tex]\[ \frac{(x - 2)(x + 4)}{2(x^2 + 4)} > 0 \][/tex]
6. Identify Critical Points:
The critical points come from the roots of the numerator. These occur at:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
Note that the denominator [tex]\( x^2 + 4 \)[/tex] never equals zero because [tex]\( x^2 + 4 > 0 \)[/tex] for all real [tex]\( x \)[/tex].
7. Test Intervals Around Critical Points:
Let's determine the sign of the expression [tex]\( \frac{(x - 2)(x + 4)}{2(x^2 + 4)} \)[/tex] in the intervals determined by our critical points: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
- For [tex]\( x \in (-\infty, -4) \)[/tex]:
Both [tex]\( (x - 2) \)[/tex] and [tex]\( (x + 4) \)[/tex] are negative, making the numerator positive.
- For [tex]\( x \in (-4, 2) \)[/tex]:
[tex]\( (x - 2) \)[/tex] is negative, [tex]\( (x + 4) \)[/tex] is positive, making the numerator negative.
- For [tex]\( x \in (2, \infty) \)[/tex]:
Both [tex]\( (x - 2) \)[/tex] and [tex]\( (x + 4) \)[/tex] are positive, making the numerator positive.
The denominator is always positive, thus the inequality [tex]\(\frac{(x - 2)(x + 4)}{2(x^2 + 4)} > 0\)[/tex] holds for intervals where the numerator is positive.
8. Conclusion:
Therefore, the solution to the inequality [tex]\(\frac{x^2 + x - 2}{x^2 + 4} > \frac{1}{2}\)[/tex] is:
[tex]\[ x \in (-\infty, -4) \cup (2, \infty) \][/tex]
1. Rewrite the Inequality:
First, let's rewrite the given inequality:
[tex]\[ \frac{x^2 + x - 2}{x^2 + 4} > \frac{1}{2} \][/tex]
2. Combine the Fractions:
To eliminate the fraction on the right-hand side, we need a common denominator:
[tex]\[ \frac{x^2 + x - 2}{x^2 + 4} - \frac{1}{2} > 0 \][/tex]
Let's combine the fractions:
[tex]\[ \frac{2(x^2 + x - 2) - (x^2 + 4)}{2(x^2 + 4)} > 0 \][/tex]
3. Simplify the Numerator:
Simplify the numerator inside the fraction:
[tex]\[ \frac{2x^2 + 2x - 4 - x^2 - 4}{2(x^2 + 4)} > 0 \][/tex]
Combining like terms, we get:
[tex]\[ \frac{x^2 + 2x - 8}{2(x^2 + 4)} > 0 \][/tex]
4. Analyze the Simplified Inequality:
Now, we have:
[tex]\[ \frac{x^2 + 2x - 8}{2(x^2 + 4)} > 0 \][/tex]
5. Factor the Numerator:
Factorize the numerator [tex]\( x^2 + 2x - 8 \)[/tex]:
[tex]\[ x^2 + 2x - 8 = (x - 2)(x + 4) \][/tex]
So the inequality becomes:
[tex]\[ \frac{(x - 2)(x + 4)}{2(x^2 + 4)} > 0 \][/tex]
6. Identify Critical Points:
The critical points come from the roots of the numerator. These occur at:
[tex]\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \][/tex]
Note that the denominator [tex]\( x^2 + 4 \)[/tex] never equals zero because [tex]\( x^2 + 4 > 0 \)[/tex] for all real [tex]\( x \)[/tex].
7. Test Intervals Around Critical Points:
Let's determine the sign of the expression [tex]\( \frac{(x - 2)(x + 4)}{2(x^2 + 4)} \)[/tex] in the intervals determined by our critical points: [tex]\( (-\infty, -4) \)[/tex], [tex]\( (-4, 2) \)[/tex], and [tex]\( (2, \infty) \)[/tex].
- For [tex]\( x \in (-\infty, -4) \)[/tex]:
Both [tex]\( (x - 2) \)[/tex] and [tex]\( (x + 4) \)[/tex] are negative, making the numerator positive.
- For [tex]\( x \in (-4, 2) \)[/tex]:
[tex]\( (x - 2) \)[/tex] is negative, [tex]\( (x + 4) \)[/tex] is positive, making the numerator negative.
- For [tex]\( x \in (2, \infty) \)[/tex]:
Both [tex]\( (x - 2) \)[/tex] and [tex]\( (x + 4) \)[/tex] are positive, making the numerator positive.
The denominator is always positive, thus the inequality [tex]\(\frac{(x - 2)(x + 4)}{2(x^2 + 4)} > 0\)[/tex] holds for intervals where the numerator is positive.
8. Conclusion:
Therefore, the solution to the inequality [tex]\(\frac{x^2 + x - 2}{x^2 + 4} > \frac{1}{2}\)[/tex] is:
[tex]\[ x \in (-\infty, -4) \cup (2, \infty) \][/tex]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your questions deserve accurate answers. Thank you for visiting IDNLearn.com, and see you again for more solutions.
