If $\alpha$ and $\beta$ are the roots of the equation $3x^2+x-1=0$, where $\alpha>\beta$, find the value of $\frac{\alpha}{\beta}+\frac{\beta}{\alpha}$. $ \textbf{(A) }\frac{7}{9}\qquad\textbf{(B) }-\frac{7}{9}\qquad\textbf{(C) }\frac{7}{3}\qquad\textbf{(D) }-\frac{7}{3}\qquad\textbf{(E) }-\frac{1}{9} $
Problem
Source:
Tags: Vieta
qwerty137
03.06.2014 09:24
What you're trying to find is $\frac{\alpha^2+\beta^2}{\alpha\beta}$, which is $\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}$. By Vieta's, $\alpha+\beta=-\frac{1}{3}$, and $\alpha\beta=-\frac{1}{3}$ as well, so $\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{\frac{1}{9}+\frac{2}{3}}{-\frac{1}{3}}=\boxed{\textbf{(D) }-\frac{7}{3}}$.
BEHZOD_UZ
05.06.2014 11:17
Ansver: D) -7/3
Rounak_iitr
05.06.2023 14:44
By vietas relation we get $$\alpha+\beta=\frac{-1}{3}$$$$\alpha\beta=\frac{-1}{3}$$$$\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{-7}{3}$$is the correct amswer!!!! Vietas Relation!!!!