Answer :
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]
