For t-1, let _t,_t be the roots of [((t+2)^1/7-1)x^2+((t+2)^1/6-1)x+((t+2)^1/21-1)=0] If …

Question

For $t>-1$, let $\alpha_t,\beta_t$ be the roots of
$$\left((t+2)^{1/7}-1\right)x^2+\left((t+2)^{1/6}-1\right)x+\left((t+2)^{1/21}-1\right)=0$$
If $\displaystyle\lim_{t\to-1^+}\alpha_t=a$ and $\displaystyle\lim_{t\to-1^+}\beta_t=b$, then $72(a+b)^2$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 98

Accepted Answer

By Vieta's formulas, $\alpha_t+\beta_t=-\dfrac{(t+2)^{1/6}-1}{(t+2)^{1/7}-1}$. Let $y=t+2\to1^+$ as $t\to-1^+$:
$$a+b=-\lim_{y\to1}\frac{y^{1/6}-1}{y^{1/7}-1}$$

Using the standard limit $\displaystyle\lim_{y\to1}\frac{y^p-1}{y^q-1}=\frac{p}{q}$ with $p=\dfrac16,q=\dfrac17$:
$$a+b=-\frac{1/6}{1/7}=-\frac76$$

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

Solutions · 1