Answer :
To solve the inequality [tex]\( x^4 - 18x^2 \geq -64 \)[/tex]:
1. Rewrite the inequality: Start by rearranging the inequality into a standard form.
[tex]\[ x^4 - 18x^2 \geq -64 \][/tex]
Add 64 to both sides to get:
[tex]\[ x^4 - 18x^2 + 64 \geq 0 \][/tex]
2. Substitute [tex]\( y = x^2 \)[/tex]: To simplify the expressions, let [tex]\( y = x^2 \)[/tex]. The inequality becomes:
[tex]\[ y^2 - 18y + 64 \geq 0 \][/tex]
3. Solve the quadratic equation: Find the roots of the quadratic equation [tex]\( y^2 - 18y + 64 = 0 \)[/tex]. The roots can be found using the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -18 \)[/tex], and [tex]\( c = 64 \)[/tex]:
[tex]\[ y = \frac{18 \pm \sqrt{324 - 256}}{2} = \frac{18 \pm \sqrt{68}}{2} = \frac{18 \pm 2\sqrt{17}}{2} = 9 \pm \sqrt{17} \][/tex]
So, the roots are:
[tex]\[ y_1 = 9 + \sqrt{17} \quad \text{and} \quad y_2 = 9 - \sqrt{17} \][/tex]
4. Convert back to [tex]\( x \)[/tex] terms: Since [tex]\( y = x^2 \)[/tex], this translates to [tex]\( x^2 = 9 + \sqrt{17} \)[/tex] and [tex]\( x^2 = 9 - \sqrt{17} \)[/tex].
Taking square roots to solve for [tex]\( x \)[/tex], we have:
[tex]\[ x = \pm \sqrt{9 + \sqrt{17}} \quad \text{and} \quad x = \pm \sqrt{9 - \sqrt{17}} \][/tex]
5. Determine the intervals:
- If [tex]\( x \ge \sqrt{9 + \sqrt{17}} \)[/tex] or [tex]\( x \le -\sqrt{9 + \sqrt{17}} \)[/tex].
- If [tex]\( -\sqrt{9 - \sqrt{17}} \le x \le \sqrt{9 - \sqrt{17}} \)[/tex].
6. Combine the intervals: The combined intervals form the solution set for the inequality.
Given the comparison and the numerical values found earlier, the correct interpretation is:
[tex]\[ (x <= \sqrt{9 - \sqrt{17}}) \cup (-\sqrt{9 - \sqrt{17}} <= x) \cup (\sqrt{9 + \sqrt{17}} \le x < \infty) \cup (-\infty < x \le -\sqrt{9 + \sqrt{17}}) \][/tex]
This corresponds to:
[tex]\[ (-\infty, -\sqrt{9+\sqrt{17}}] \cup [-\sqrt{9-\sqrt{17}}, \sqrt{9-\sqrt{17}}] \cup [\sqrt{9+\sqrt{17}}, \infty) \][/tex]
Calculating the approximate values:
[tex]\[ \sqrt{9+\sqrt{17}} \approx 3.623, \quad \sqrt{9-\sqrt{17}} \approx 2.208 \][/tex]
Thus:
[tex]\[ (-\infty, -3.623] \cup [-2.208, 2.208] \cup [3.623, \infty) \][/tex]
Therefore, the correct solution interval for the inequality [tex]\( x^4 - 18x^2 \geq -64 \)[/tex] is:
[tex]\[ (-\infty, -3.623] \cup [-2.208, 2.208] \cup [3.623, \infty) \][/tex]
1. Rewrite the inequality: Start by rearranging the inequality into a standard form.
[tex]\[ x^4 - 18x^2 \geq -64 \][/tex]
Add 64 to both sides to get:
[tex]\[ x^4 - 18x^2 + 64 \geq 0 \][/tex]
2. Substitute [tex]\( y = x^2 \)[/tex]: To simplify the expressions, let [tex]\( y = x^2 \)[/tex]. The inequality becomes:
[tex]\[ y^2 - 18y + 64 \geq 0 \][/tex]
3. Solve the quadratic equation: Find the roots of the quadratic equation [tex]\( y^2 - 18y + 64 = 0 \)[/tex]. The roots can be found using the quadratic formula, [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -18 \)[/tex], and [tex]\( c = 64 \)[/tex]:
[tex]\[ y = \frac{18 \pm \sqrt{324 - 256}}{2} = \frac{18 \pm \sqrt{68}}{2} = \frac{18 \pm 2\sqrt{17}}{2} = 9 \pm \sqrt{17} \][/tex]
So, the roots are:
[tex]\[ y_1 = 9 + \sqrt{17} \quad \text{and} \quad y_2 = 9 - \sqrt{17} \][/tex]
4. Convert back to [tex]\( x \)[/tex] terms: Since [tex]\( y = x^2 \)[/tex], this translates to [tex]\( x^2 = 9 + \sqrt{17} \)[/tex] and [tex]\( x^2 = 9 - \sqrt{17} \)[/tex].
Taking square roots to solve for [tex]\( x \)[/tex], we have:
[tex]\[ x = \pm \sqrt{9 + \sqrt{17}} \quad \text{and} \quad x = \pm \sqrt{9 - \sqrt{17}} \][/tex]
5. Determine the intervals:
- If [tex]\( x \ge \sqrt{9 + \sqrt{17}} \)[/tex] or [tex]\( x \le -\sqrt{9 + \sqrt{17}} \)[/tex].
- If [tex]\( -\sqrt{9 - \sqrt{17}} \le x \le \sqrt{9 - \sqrt{17}} \)[/tex].
6. Combine the intervals: The combined intervals form the solution set for the inequality.
Given the comparison and the numerical values found earlier, the correct interpretation is:
[tex]\[ (x <= \sqrt{9 - \sqrt{17}}) \cup (-\sqrt{9 - \sqrt{17}} <= x) \cup (\sqrt{9 + \sqrt{17}} \le x < \infty) \cup (-\infty < x \le -\sqrt{9 + \sqrt{17}}) \][/tex]
This corresponds to:
[tex]\[ (-\infty, -\sqrt{9+\sqrt{17}}] \cup [-\sqrt{9-\sqrt{17}}, \sqrt{9-\sqrt{17}}] \cup [\sqrt{9+\sqrt{17}}, \infty) \][/tex]
Calculating the approximate values:
[tex]\[ \sqrt{9+\sqrt{17}} \approx 3.623, \quad \sqrt{9-\sqrt{17}} \approx 2.208 \][/tex]
Thus:
[tex]\[ (-\infty, -3.623] \cup [-2.208, 2.208] \cup [3.623, \infty) \][/tex]
Therefore, the correct solution interval for the inequality [tex]\( x^4 - 18x^2 \geq -64 \)[/tex] is:
[tex]\[ (-\infty, -3.623] \cup [-2.208, 2.208] \cup [3.623, \infty) \][/tex]
