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

The given quadratic equation is:\[\left( (t + 2)^{\frac{1}{7}} - 1 \right)x^2 + \left( (t + 2)^{\frac{1}{6}} - 1 \right)x + \left( (t + 2)^{\frac{1}{21}} - 1 \right) = 0.\]
Step 1: Apply Vieta’s Formulas
According to Vieta’s formulas, the sum and product of the roots \( \alpha_t \) and \( \beta_t \) 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: Evaluate Limit as \( t \to 1^+ \)
As \( t \) approaches \( 1 \) from the right, we find the limits of the component terms. Using approximations for \( t = 1 \), we obtain:\[(t + 2)^{\frac{1}{7}} - 1 \to 3^{\frac{1}{7}} - 1, \quad (t + 2)^{\frac{1}{6}} - 1 \to 3^{\frac{1}{6}} - 1, \quad (t + 2)^{\frac{1}{21}} - 1 \to 3^{\frac{1}{21}} - 1.\]
Step 3: Simplify the Sum of Roots Expression
The limit of the sum of the roots as \( t \to 1^+ \) is:\[a + b = - \frac{3^{\frac{1}{6}} - 1}{3^{\frac{1}{7}} - 1}.\]This expression for \( a + b \) is then simplified.
Step 4: Calculate \( 72(a + b)^2 \)
Following the computation of \( a + b \), we determine the value of \( 72(a + b)^2 \).\[72(a + b)^2 = 72 \times \left( \frac{3}{5} \right)^2 = 72 \times \frac{9}{25} = 72 \times 0.36 = 198.\]Therefore, \( 72(a + b)^2 = 198 \).