A) \[n:m:\ell \]
B) \[\ell :m:n\]
C) \[m:n:\ell \]
D) \[n:\ell :m\]
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 \[\ell {{x}^{2}}+mx+n=0\,\,\,are\,\,\frac{1}{\alpha },\,\,\frac{1}{\beta }\] 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 \[\frac{a}{n}=-\frac{b}{m}\,=\frac{c}{\ell }\,\,\therefore \,\,a:b:c=n:m:\ell \]
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