Question
If the roots of the equation $$ax^2 – bx + c = 0$$ are $$\alpha , \beta$$ then the roots of the equation $${b^2}c{x^2} – a{b^2}x + {a^3} = 0$$ are
A.
$$\frac{1}{{{\alpha ^3} + \alpha \beta }},\frac{1}{{{\beta ^3} + \alpha \beta }}$$
B.
$$\frac{1}{{{\alpha ^2} + \alpha \beta }},\frac{1}{{{\beta ^2} + \alpha \beta }}$$
C.
$$\frac{1}{{{\alpha ^4} + \alpha \beta }},\frac{1}{{{\beta ^4} + \alpha \beta }}$$
D.
None of these
Answer :
$$\frac{1}{{{\alpha ^2} + \alpha \beta }},\frac{1}{{{\beta ^2} + \alpha \beta }}$$
Solution :
Multiplying the second equation by $$\frac{c}{{{a^3}}},$$
we get, $$\frac{{{b^2}{c^2}}}{{{a^3}}}{x^2} - \frac{{{b^2}c}}{{{a^2}}}x + c = 0$$
$$\eqalign{
& \Rightarrow a{\left( {\frac{{bc}}{{{a^2}}}x} \right)^2} - b\left( {\frac{{bc}}{{{a^2}}}} \right)x + c = 0 \cr
& \Rightarrow \frac{{bc}}{{{a^2}}}x = \alpha ,\beta \cr
& \Rightarrow \left( {\alpha + \beta } \right)\alpha \beta x = \alpha ,\beta \cr
& \Rightarrow x = \frac{1}{{\left( {\alpha + \beta } \right)\alpha }},\frac{1}{{\left( {\alpha + \beta } \right)\beta }} \cr} $$