Complex Numbers And Quadratic Equations question 703
Question: If the roots of $ ax^{2}+bx+c=0 $ are the reciprocals of those of $ \ell x^{2}+mx+n=0 $ then $ a:b:c= $
Options:
A) $ n:m:\ell $
B) $ \ell :m:n $
C) $ m:n:\ell $
D) $ n:\ell :m $
Show Answer
Answer:
Correct Answer: A
Solution:
If $ \alpha ,\beta $ be the roots then $ \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a} $ Now the roots of $ \ell x^{2}+mx+n=0are\frac{1}{\alpha },\frac{1}{\beta } $ are
$ \therefore \frac{1}{\alpha }+\frac{1}{\beta }=-\frac{m}{\ell }and\frac{1}{\alpha }.\frac{1}{\beta }=\frac{n}{\ell } $ $ \frac{\alpha +\beta }{\alpha \beta }=-\frac{m}{\ell }and\frac{a}{c}=\frac{n}{\ell } $ $ -\frac{b}{c}=-\frac{m}{\ell }and\frac{a}{c}=\frac{n}{\ell } $ or $ \frac{a}{n}=-\frac{b}{m}=\frac{c}{\ell }\therefore a:b:c=n:m:\ell $
BETA
