For t > –1, let αt and βt be the roots of the equation ((t+2)^(1/7)-1)x^2 + ((t+2)^(1/6)-1)x + ((t+2)^(1/21)-1) = 0.

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 \(\lim\limits _{t \rightarrow-1^{+}} \alpha_{t}=a\) and \(\lim\limits _{t \rightarrow-+^{+}} \beta_{t}=b,\) then \(72(a+b)^{2}\) is equal to _____.

Answer 1

Answer is: 98 

\(a+b=\lim\limits _{t \rightarrow-1^{+}}(\alpha+\beta)=\lim\limits _{t \rightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1}\)

let \(\mathrm{t}+2=\mathrm{y}\)

\(a+b=\lim\limits _{y \rightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6}\)

\(72(a+b)^{2}=72 \frac{49}{36}=98\)