Complex Numbers And Quadratic Equations question 652
Question: If $ \alpha $ and $ \beta $ be the values of x in $ m^{2}(x^{2}-x)+2mx+3=0 $ and $ m_1 $ and $ m_2 $ be two values of m for which $ \alpha $ and $ \beta $ are connected by the relation $ \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{3}. $ Then the value of $ \frac{m_1^{2}}{m_2}+\frac{m_2^{2}}{m_1} $ is
Options:
A) 6
B) 68
C) $ \frac{3}{68} $
D) $ -\frac{68}{3} $
Show Answer
Answer:
Correct Answer: D
Solution:
The given equation is $ {m^{2}}x^{2}+( 2m-m^{2} )x+3=0 $
$ \therefore \alpha +\beta =-\frac{2m-m^{2}}{m^{2}}=\frac{m-2}{m}and\alpha \beta =\frac{3}{m^{2}} $ Now $ \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{4}{3}\Rightarrow \frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }=\frac{4}{3}\Rightarrow \frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }=\frac{4}{3} $ Substituting the values, we get $ \frac{{{( \frac{m-2}{m} )}^{2}}-2.\frac{3}{m^{2}}}{\frac{3}{m^{2}}}=\frac{4}{3} $
$ \Rightarrow \frac{m^{2}-4m+4-6}{3}=\frac{4}{3}\Rightarrow m^{2}-4m-6=0 $ $ {m_1}andm_2 $ are roots of this equation, therefore $ {m_1}+m_2=4andm_1m_2=-6 $ The given expression is, $ \frac{m_1^{2}}{m_2}+\frac{m_2^{2}}{m_1}=\frac{m_1^{3}+m_2^{3}}{m_1m_2} $ $ =\frac{{{(m_1+m_2)}^{3}}-3m_1m_2(m_1+m_2)}{m_1m_2} $ $ =\frac{{{(4)}^{3}}-3.(-6).(4)}{-6}=-\frac{68}{3} $
BETA
