Complex Numbers And Quadratic Equations question 592
Question: Sum of roots is $ -1 $ and sum of their reciprocals is $ \frac{1}{6} $ , then equation is [Karnataka CET 1998]
Options:
A) $ x^{2}+x-6=0 $
B) $ x^{2}-x+6=0 $
C) $ 6x^{2}+x+1=0 $
D) $ x^{2}-6x+1=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let roots are $ \alpha $ and $ \beta $ . Given $ \alpha +\beta =-1 $ $ \frac{1}{\alpha }+\frac{1}{\beta }=\frac{1}{6}\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=\frac{1}{6}\Rightarrow \alpha \beta =-6 $ Hence the equation, $ x^{2}-(\alpha +\beta )x+\alpha \beta =0 $
Þ $ x^{2}+x-6=0 $
BETA
