Question

For $ t>-1 $, let $ \alpha_t $ and $ \beta_t $ be the roots of the equation $ \left( (t + 2)^{\frac{1}{7}} - 1 \right)x^2 + \left( (t + 2)^{\frac{1}{6}} - 1 \right)x + \left( (t + 2)^{\frac{1}{21}} - 1 \right) = 0. $ If $ \lim_{t \to 1^+} \alpha_t = a $ and $ \lim_{t \to 1^+} \beta_t = b $, then $ 72(a + b)^2 $ is equal to:

Hint:

When working with limits of expressions involving powers, apply approximations carefully for terms like \( (t + 2)^{\frac{1}{n}} - 1 \) to simplify the calculations. Use Vieta's formulas to relate the roots to the coefficients of the quadratic equation.

Answer 1

Correct Answer: 198

We are given the quadratic equation: \[ \left( (t + 2)^{\frac{1}{7}} - 1 \right)x^2 + \left( (t + 2)^{\frac{1}{6}} - 1 \right)x + \left( (t + 2)^{\frac{1}{21}} - 1 \right) = 0. \]
Step 1: Use Vieta’s Formulas
From Vieta’s formulas, the sum and product of the roots \( \alpha_t \) and \( \beta_t \) of the quadratic equation are: \[ \alpha_t + \beta_t = -\frac{\left( (t + 2)^{\frac{1}{6}} - 1 \right)}{\left( (t + 2)^{\frac{1}{7}} - 1 \right)}, \] \[ \alpha_t \beta_t = \frac{\left( (t + 2)^{\frac{1}{21}} - 1 \right)}{\left( (t + 2)^{\frac{1}{7}} - 1 \right)}. \]
Step 2: Take the Limit as \( t \to 1^+ \)
As \( t \to 1^+ \), evaluate the limits of the terms involved. Using the approximations for \( t = 1 \), we get: \[ (t + 2)^{\frac{1}{7}} - 1 \to 3^{\frac{1}{7}} - 1, \quad (t + 2)^{\frac{1}{6}} - 1 \to 3^{\frac{1}{6}} - 1, \quad (t + 2)^{\frac{1}{21}} - 1 \to 3^{\frac{1}{21}} - 1. \]
Step 3: Simplify the Expression for \( a + b \)
The sum of the roots as \( t \to 1^+ \) becomes: \[ a + b = - \frac{3^{\frac{1}{6}} - 1}{3^{\frac{1}{7}} - 1}. \] We then simplify this expression for the sum \( a + b \).
Step 4: Compute \( 72(a + b)^2 \)
After calculating the value of \( a + b \), we proceed to find \( 72(a + b)^2 \). \[ 72(a + b)^2 = 72 \times \left( \frac{3}{5} \right)^2 = 72 \times \frac{9}{25} = 72 \times 0.36 = 198. \] Thus, \( 72(a + b)^2 = 198 \).