Complex Numbers And Quadratic Equations question 594
Question: If a, b are roots of $ x^{2}-3x+1=0, $ then the equation whose roots are $ \frac{1}{\alpha -2},\frac{1}{\beta -2} $ is [RPET 1999]
Options:
A) $ x^{2}+x-1=0 $
B) $ x^{2}+x+1=0 $
C) $ x^{2}-x-1=0 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \alpha ,\beta $ are the roots of the equation $ x^{2}-3x+1=0 $
$ \therefore \alpha +\beta =3 $ and $ \alpha \beta =1 $ $ S=\frac{1}{\alpha -2}+\frac{1}{\beta -2}=\frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4} $ $ =\frac{3-4}{1-2.3+4}=1 $ and $ P=\frac{1}{(\alpha -2)(\beta -2)}=\frac{1}{\alpha \beta -2(\alpha +\beta )+4}=-1 $ Hence the equation whose roots are $ \frac{1}{\alpha -2} $ and $ \frac{1}{\beta -2} $ are $ x^{2}-Sx+P=0\Rightarrow x^{2}-x-1=0 $ .
BETA
