Discussion Forum

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>-1,

 Then , $\lim_{a \rightarrow {0^{+}}}\alpha(a)$ and $\lim_{a \rightarrow {0^{+}}}\beta(a)$ are 

Answer:

Explanation:

Concept Involved To make the quadratic into the simple form we should eliminate radical sign 

Description of Situation As forgiven equation when  $a\rightarrow 0$  equation reduces to identity in x.

i.e, $ax^{2}+bx+c=0$ for all $x \epsilon  R$ or a=b=c $\rightarrow 0$

Thus, first we should make above equation indepenrlent from coefficients as 0.

 Sol. Let $(a+1=t^{6}$ Thus, when $a \rightarrow 0,t \rightarrow 1$

 $\therefore$  $(t^{2}-1)x^{2}+(t^{3}-1)x+(t-1)=0$

$\Rightarrow$  $(t-1){(t+1)x^{2}+(t^{2}+t+1)x+1}=0$

 As $t \rightarrow 1$ 

                     $2x^{2}+3x+1=0$

 $2x^{2}+2x+x+1=0$

$\Rightarrow$   $(2x+1) (x+1)=0$

 Thus, x=-1, -1/2

 or  $\lim_{a \rightarrow {0^{+}}}\alpha(a)=-\frac{1}{2}$

 and   $\lim_{a \rightarrow {0^{+}}}\beta(a)=-1$