Complex Numbers And Quadratic Equations question 530
Question: If $ \alpha ,\ \beta $ are the roots of the equation $ ax^{2}+bx+c=0 $ , then $ \frac{\alpha }{a\beta +b}+\frac{\beta }{a\alpha +b}= $
Options:
A) $ \frac{2}{a} $
B) $ \frac{2}{b} $
C) $ \frac{2}{c} $
D) $ -\frac{2}{a} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a} $ and $ {{\alpha }^{2}}+{{\beta }^{2}}=\frac{(b^{2}-2ac)}{a^{2}} $ Now $ \frac{\alpha }{ \begin{aligned} & a\beta +b \\ & \\ \end{aligned}}+\frac{\beta }{a\alpha +b} $ $ =\frac{\alpha (a\alpha +b)+\beta (a\beta +b)}{(a\beta +b)(a\alpha +b)} $ $ =\frac{a({{\alpha }^{2}}+{{\beta }^{2}})+b(\alpha +\beta )}{\alpha \beta a^{2}+ab(\alpha +\beta )+b^{2}}=\frac{a\frac{(b^{2}-2ac)}{a^{2}}+b( -\frac{b}{a} )}{( \frac{c}{a} )a^{2}+ab( -\frac{b}{a} )+b^{2}} $ $ =\frac{b^{2}-ac-b^{2}}{a^{2}c-ab^{2}+ab^{2}}=\frac{-2ac}{a^{2}c}=-\frac{2}{a} $ .
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