Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1)…
Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^2+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a\gt-1$ then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{a \rightarrow 0^{+}} \beta(a)$ respectively are
1 and $-\frac{5}{2}$
-1 and $-\frac{1}{2}$
2 and $-\frac{7}{2}$
3 and $-\frac{9}{2}$
Solution
Let $A=1+a$
$\therefore \quad$ When $\mathrm{a} \rightarrow 0^{+}, \mathrm{A} \rightarrow 1^{+}$
$\therefore \quad$ Given function is written as
$\begin{aligned}
& \left(A^{\frac{1}{3}}-1\right) x^2+\left(A^{\frac{1}{2}}-1\right) x+\left(A^{\frac{1}{6}}-1\right)=0 \\
\therefore & \left(\frac{A^{\frac{1}{3}}-1}{A-1}\right) x^2+\left(\frac{A^{\frac{1}{2}}-1}{A-1}\right) x+\left(\frac{A^{\frac{1}{6}}-1}{A-1}\right)=0
\end{aligned}$ Taking $\lim _{x \rightarrow 0^{+}}$on both sides, we get
$\begin{array}{ll}
\therefore & \frac{1}{3} x^2+\frac{1}{2} x+\frac{1}{6}=0 \\
\therefore & 2 x^2+3 x+1=0 \\
\therefore & x=-1 \text { or } \frac{-1}{2}
\end{array}$
i.e. $\lim _{x \rightarrow 0^{+}} \alpha(a)=-1$ and $\lim _{x \rightarrow 0^{+}} \beta(a)=\frac{-1}{2}$