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The problem is asking to find the interval of time over which the temperature \( T(t) \), given by the function

\[
T(t) = \frac{8t}{t^2 + 1} + 98.6
\]

is greater than 100°F. 

### Step-by-step approach:

1. **Set up the inequality**: To find when the temperature is greater than 100°F, we set up the inequality:

   \[
   \frac{8t}{t^2 + 1} + 98.6 > 100
   \]

2. **Isolate the fraction**: Subtract 98.6 from both sides:

   \[
   \frac{8t}{t^2 + 1} > 100 - 98.6
   \]

   \[
   \frac{8t}{t^2 + 1} > 1.4
   \]

3. **Multiply both sides by \( t^2 + 1 \)** to eliminate the denominator (since \( t^2 + 1 > 0 \) for all real \( t \)):

   \[
   8t > 1.4(t^2 + 1)
   \]

4. **Expand the right-hand side**:

   \[
   8t > 1.4t^2 + 1.4
   \]

5. **Rearrange the inequality**:

   \[
   0 > 1.4t^2 - 8t + 1.4
   \]

   or

   \[
   1.4t^2 - 8t + 1.4 < 0
   \]

6. **Solve the quadratic inequality**: First, solve the corresponding quadratic equation:

   \[
   1.4t^2 - 8t + 1.4 = 0
   \]

   Use the quadratic formula:

   \[
   t = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1.4)(1.4)}}{2(1.4)}
   \]

   Simplifying:

   \[
   t = \frac{8 \pm \sqrt{64 - 7.84}}{2.8}
   \]

   \[
   t = \frac{8 \pm \sqrt{56.16}}{2.8}
   \]

   \[
   t = \frac{8 \pm 7.49}{2.8}
   \]

   So the two solutions are:

   \[
   t_1 = \frac{8 - 7.49}{2.8} \approx 0.18, \quad t_2 = \frac{8 + 7.49}{2.8} \approx 5.52
   \]

7. **Interpret the solution**: The quadratic inequality \( 1.4t^2 - 8t + 1.4 < 0 \) is satisfied between the roots. So, the temperature is over 100°F in the interval:

   \[
   t \in (0.18, 5.52)
   \]

### Final answer: The temperature was over 100°F between approximately **0.18 hours** and **5.52 hours**.

Would you like more details on any step or have additional questions?

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### Related questions:

1. What is the significance of solving quadratic inequalities in real-world applications?
2. How does the quadratic formula help in solving non-linear inequalities?
3. Can this model for temperature change be applied to other types of illnesses?
4. How would you graph this function to visualize the temperature changes?
5. What are the methods to solve inequalities besides using the quadratic formula?

### Tip:
When solving quadratic inequalities, always check the sign of the quadratic expression in the intervals determined by the roots to verify the solution's correctness.

The temperature T(t), in degrees Fahrenheit, of a person during an illness is given by the function T(t) = 8t / (t^2 + 1) + 98.6, where t = time, in hours. Find the interval over which the temperature was over 100°F.

Learn how to solve the inequality for temperature T(t) > 100°F using a quadratic equation. This problem involves key algebraic concepts such as inequalities, quadratic equations, and solving for intervals. Suitable for students in Grades 10-12.

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