For $t \gt -1$, 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 $\lim _{t \rightarrow-1^{+}} \alpha_t=a$ and $\lim _{t \rightarrow-1^{+}} \beta_t=b$, then $72(a+b)^2$ is equal to ________.

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Aicharya · Accepted Answer

$\begin{aligned} & a+b=\lim _{t ightarrow-1^{+}}(\alpha+\beta)=\lim _{t ightarrow-1^{+}}-\frac{(t+2)^{\frac{1}{6}}-1}{(t+2)^{\frac{1}{7}}-1} \ & 	ext { let } t+2=y \ & a+b=\lim _{y ightarrow 1^{+}} \frac{y^{1 / 6}-1}{y^{1 / 7}-1}=\frac{7}{6} \ & 72(a+b)^2=72 \frac{49}{36}=98\end{aligned}$

For $t \gt -1$, let $\alpha_t$ and $\beta_t$ be the roots of the equation…

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