Complex Numbers And Quadratic Equations question 596
Question: If a, b are the roots of $ 9x^{2}+6x+1=0, $ then the equation with the roots $ \frac{1}{\alpha },\frac{1}{\beta } $ is [EAMCET 2000]
Options:
A) $ 2x^{2}+3x+18=0 $
B) $ x^{2}+6x-9=0 $
C) $ x^{2}+6x+9=0 $
D) $ x^{2}-6x+9=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation is $ 9x^{2}+6x+1=0 $
$ \Rightarrow \alpha +\beta =\frac{-6}{9}=\frac{-2}{3} $ and $ \alpha \beta =1/9 $
$ \therefore \alpha -\beta =\sqrt{{{(\alpha +\beta )}^{2}}-4\alpha \beta } $ $ =\sqrt{\frac{4}{9}-4.\frac{1}{9}}=0 $
$ \Rightarrow \alpha =\frac{-1}{3},\beta =\frac{-1}{3} $ \ Equation $ x^{2}-(\alpha +\beta )x+\alpha \beta =0 $
$ \Rightarrow x^{2}+6x+9=0 $ .
BETA
