Key Concepts and Formulas

• Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. If $a, b, c$ are the side lengths, then $a + b > c$, $a + c > b$, and $b + c > a$.
• Quadratic Formula: The roots of the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Discriminant of a Quadratic: The discriminant of the quadratic $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. If $a > 0$ and $\Delta < 0$, then $ax^2 + bx + c > 0$ for all real $x$.

Step-by-Step Solution

Step 1: State the Triangle Inequality Theorem for the given sides

We are given the side lengths $5, 5r, 5r^2$. Applying the Triangle Inequality Theorem, we have: \begin{enumerate} \item $5 + 5r > 5r^2$ \item $5 + 5r^2 > 5r$ \item $5r + 5r^2 > 5$ \end{enumerate} Since the side lengths must be positive, we also have $r > 0$. We will consider this condition when we find the intersection of the solution sets.

Step 2: Simplify the inequalities

Dividing each inequality by 5 (since 5 is positive, the inequality signs remain the same), we get: \begin{enumerate} \item $1 + r > r^2 \implies r^2 - r - 1 < 0$ \item $1 + r^2 > r \implies r^2 - r + 1 > 0$ \item $r + r^2 > 1 \implies r^2 + r - 1 > 0$ \end{enumerate}

Step 3: Solve the first inequality: $r^2 - r - 1 < 0$

To solve $r^2 - r - 1 < 0$, we first find the roots of the quadratic equation $r^2 - r - 1 = 0$ using the quadratic formula: $r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$ The roots are $r_1 = \frac{1 - \sqrt{5}}{2}$ and $r_2 = \frac{1 + \sqrt{5}}{2}$. Since the coefficient of $r^2$ is positive, the parabola opens upwards. The inequality $r^2 - r - 1 < 0$ holds for $r$ between the roots. Thus, $\frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2}$

Step 4: Solve the second inequality: $r^2 - r + 1 > 0$

Consider the quadratic $r^2 - r + 1$. The discriminant is $\Delta = (-1)^2 - 4(1)(1) = 1 - 4 = -3$. Since $\Delta < 0$ and the coefficient of $r^2$ is positive, the quadratic is always positive for all real values of $r$. Thus, $r^2 - r + 1 > 0$ for all $r \in \mathbb{R}$.

Step 5: Solve the third inequality: $r^2 + r - 1 > 0$

To solve $r^2 + r - 1 > 0$, we first find the roots of the quadratic equation $r^2 + r - 1 = 0$ using the quadratic formula: $r = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}$ The roots are $r_3 = \frac{-1 - \sqrt{5}}{2}$ and $r_4 = \frac{-1 + \sqrt{5}}{2}$. Since the coefficient of $r^2$ is positive, the parabola opens upwards. The inequality $r^2 + r - 1 > 0$ holds for $r$ outside the roots. Thus, $r < \frac{-1 - \sqrt{5}}{2} \quad \text{or} \quad r > \frac{-1 + \sqrt{5}}{2}$

Step 6: Combine all conditions

We need to find the intersection of the following intervals: \begin{enumerate} \item $r > 0$ \item $\frac{1 - \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2}$ \item $r \in \mathbb{R}$ \item $r < \frac{-1 - \sqrt{5}}{2}$ or $r > \frac{-1 + \sqrt{5}}{2}$ \end{enumerate} Since $r > 0$, we can ignore the parts of intervals 2 and 4 where $r < 0$. We have: \begin{enumerate} \item $r > 0$ \item $0 < r < \frac{1 + \sqrt{5}}{2}$ \item $r \in \mathbb{R}$ \item $r > \frac{-1 + \sqrt{5}}{2}$ \end{enumerate} Combining these, we get $\frac{-1 + \sqrt{5}}{2} < r < \frac{1 + \sqrt{5}}{2}$.

Let $A = \frac{-1 + \sqrt{5}}{2}$ and $B = \frac{1 + \sqrt{5}}{2}$. Then $r \in (A, B)$. Approximately, $A \approx \frac{-1 + 2.236}{2} \approx 0.618$ and $B \approx \frac{1 + 2.236}{2} \approx 1.618$.

Step 7: Check the given options

We need to find the value of $r$ that is not in the interval $(A, B)$. \begin{enumerate} \item $r = \frac{7}{4} = 1.75$. Since $1.75 > 1.618$, $\frac{7}{4}$ is not in the interval. \item $r = \frac{5}{4} = 1.25$. Since $0.618 < 1.25 < 1.618$, $\frac{5}{4}$ is in the interval. \item $r = \frac{3}{4} = 0.75$. Since $0.618 < 0.75 < 1.618$, $\frac{3}{4}$ is in the interval. \item $r = \frac{3}{2} = 1.5$. Since $0.618 < 1.5 < 1.618$, $\frac{3}{2}$ is in the interval. \end{enumerate} Therefore, $r$ cannot be equal to $\frac{7}{4}$.

Common Mistakes & Tips

• Remember to consider the condition that side lengths must be positive, which implies $r > 0$.
• When solving quadratic inequalities, pay attention to the sign of the leading coefficient to determine the interval where the inequality holds.
• Approximating the roots of the quadratic equations can help in comparing the given options.

Summary

We used the Triangle Inequality Theorem to establish three inequalities involving $r$. We solved each inequality and found the intersection of their solution sets, along with the condition $r > 0$. By checking the given options, we determined that $r$ cannot be equal to $\frac{7}{4}$.

Final Answer

The final answer is \boxed{\frac{7}{4}}, which corresponds to option (A).