IDNLearn.com is your go-to resource for finding precise and accurate answers. Find accurate and detailed answers to your questions from our experienced and dedicated community members.
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 using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
