For t -1,let alpha_t and beta_t be the root | vedclass

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(C) Let $y = (t+2)^{\frac{1}{42}}$. As $t \rightarrow -1^{+}$,$y \rightarrow 1^{+}$.Then $(t+2)^{\frac{1}{7}} = y^6$,$(t+2)^{\frac{1}{6}} = y^7$,and $

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(C) Let $y = (t+2)^{\frac{1}{42}}$. As $t ightarrow -1^{+}$,$y ightarrow 1^{+}$.Then $(t+2)^{\frac{1}{7}} = y^6$,$(t+2)^{\frac{1}{6}} = y^7$,and $(t+2)^{\frac{1}{21}} = y^2$.The equation becomes $(y^6-1)x^2 + (y^7-1)x + (y^2-1) = 0$.Dividing by $(y-1)$,we get $\frac{y^6-1}{y-1}x^2 + \frac{y^7-1}{y-1}x + \frac{y^2-1}{y-1} = 0$.Taking the limit as $y ightarrow 1$,we get $6x^2 + 7x + 2 = 0$.The sum of the roots $a+b = -\frac{7}{6}$.Thus,$72(a+b)^2 = 72 	imes \left(-\frac{7}{6}ight)^2 = 72 	imes \frac{49}{36} = 2 	imes 49 = 98$.

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