Key Concepts and Formulas

• Sum of Roots of a Quadratic Equation: For a quadratic equation of the form $Ax^2 + Bx + C = 0$, the sum of the roots is given by $-\frac{B}{A}$.
• Limits of the form $\frac{0}{0}$: When evaluating limits that result in an indeterminate form like $\frac{0}{0}$, L'Hôpital's Rule or algebraic manipulation using standard limit forms can be applied.
• Standard Limit Form: The limit $\lim_{x \to c} \frac{x^n - c^n}{x - c} = nc^{n-1}$ can be used, or equivalently, $\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n$.

Step-by-Step Solution

Step 1: Identify the sum of the roots ($\alpha_t + \beta_t$) in terms of the coefficients of the quadratic equation. The given quadratic equation is: $\left((\mathrm{t}+2)^{1 / 7}-1\right) x^2+\left((\mathrm{t}+2)^{1 / 6}-1\right) x+\left((\mathrm{t}+2)^{1 / 21}-1\right)=0$ Here, $A = (\mathrm{t}+2)^{1 / 7}-1$, $B = (\mathrm{t}+2)^{1 / 6}-1$, and $C = (\mathrm{t}+2)^{1 / 21}-1$. The sum of the roots, $\alpha_t + \beta_t$, is given by $-\frac{B}{A}$. Therefore, $\alpha_t + \beta_t = -\frac{(\mathrm{t}+2)^{1 / 6}-1}{(\mathrm{t}+2)^{1 / 7}-1}$.

Step 2: Evaluate the limit of the sum of the roots as $t \to -1^+$. We are given that $a = \lim _{t \rightarrow-1^{+}} \alpha_{t}$ and $b = \lim _{t \rightarrow-1^{+}} \beta_{t}$. The sum of the limits is $a+b = \lim _{t \rightarrow-1^{+}} \alpha_{t} + \lim _{t \rightarrow-1^{+}} \beta_{t}$. Assuming the limits of individual roots exist, we can write: $a+b = \lim _{t \rightarrow-1^{+}}(\alpha_t + \beta_t) = \lim _{t \rightarrow-1^{+}} -\frac{(\mathrm{t}+2)^{1 / 6}-1}{(\mathrm{t}+2)^{1 / 7}-1}$

Step 3: Simplify the limit expression by a substitution and algebraic manipulation. Let $y = t+2$. As $t \rightarrow -1^{+}$, we have $y \rightarrow (-1+2)^{+} = 1^{+}$. Substituting $y$ into the limit expression: $a+b = \lim _{y \rightarrow 1^{+}} -\frac{y^{1 / 6}-1}{y^{1 / 7}-1}$ To evaluate this limit, we can rewrite it in a form suitable for the standard limit formula $\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n$. We can multiply the numerator and denominator by $(y-1)$: $a+b = -\lim _{y \rightarrow 1^{+}} \frac{\frac{y^{1 / 6}-1}{y-1}}{\frac{y^{1 / 7}-1}{y-1}}$ Now, we can apply the standard limit formula to the numerator and the denominator separately. For the numerator, $\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y-1}$ with $n = 1/6$ is equal to $1/6$. For the denominator, $\lim _{y \rightarrow 1^{+}} \frac{y^{1 / 7}-1}{y-1}$ with $n = 1/7$ is equal to $1/7$. Therefore, $a+b = -\frac{1/6}{1/7} = -\frac{7}{6}$

Step 4: Calculate the value of $72(a+b)^2$. We have found that $a+b = -\frac{7}{6}$. Now, we need to calculate $72(a+b)^2$: $72(a+b)^2 = 72 \left(-\frac{7}{6}\right)^2$ $72(a+b)^2 = 72 \left(\frac{49}{36}\right)$ $72(a+b)^2 = \frac{72}{36} \times 49$ $72(a+b)^2 = 2 \times 49$ $72(a+b)^2 = 98$

Common Mistakes & Tips

• L'Hôpital's Rule Application: Be careful when applying L'Hôpital's Rule to ensure the limit is indeed in the $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form. In this case, direct substitution of $y=1$ into the numerator and denominator gives $1^{1/6}-1=0$ and $1^{1/7}-1=0$, confirming the $\frac{0}{0}$ form.
• Sign Errors: The sum of roots formula has a negative sign ($-\frac{B}{A}$). Ensure this sign is carried through the calculation.
• Standard Limit Form: Recognize that the limit form $\lim_{y \to 1} \frac{y^n - 1}{y^m - 1}$ can be directly evaluated as $\frac{n}{m}$ by dividing the numerator and denominator by $(y-1)$ and applying $\lim_{y \to 1} \frac{y^k - 1}{y - 1} = k$.

Summary

The problem requires finding the value of $72(a+b)^2$, where $a$ and $b$ are the limits of the roots of a given quadratic equation as $t \to -1^+$. We first used the sum of roots formula for a quadratic equation to express $\alpha_t + \beta_t$ in terms of the coefficients. Then, we evaluated the limit of this sum as $t \to -1^+$ by making a substitution and utilizing the standard limit form $\lim_{x \to 1} \frac{x^n - 1}{x - 1} = n$. This allowed us to find the value of $a+b$, which was then squared and multiplied by 72 to obtain the final answer.

The final answer is $\boxed{98}$.