Complex Numbers And Quadratic Equations question 194
Question: The value of ‘a’ for which one root of the quadratic equation $ (a^{2}-5a+3)x^{2}+(3a-1)x+2=0 $ is twice as large as the other, is [AIEEE 2003]
Options:
A) $ \frac{2}{3} $
B) $ -\frac{2}{3} $
C) $ \frac{1}{3} $
D) $ -\frac{1}{3} $
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Answer:
Correct Answer: A
Solution:
Let the roots are a and 2a
Þ $ \alpha +2\alpha =\frac{1-3a}{a^{2}-5a+3} $ and $ \alpha .2\alpha =\frac{2}{a^{2}-5a+3} $
Þ $ 2[ \frac{1}{9}\frac{{{(1-3a)}^{2}}}{{{(a^{2}-5a+3)}^{2}}} ]=\frac{2}{a^{2}-5a+3} $
Þ $ \frac{{{(1-3a)}^{2}}}{(a^{2}-5a+3)}=9 $
$ \Rightarrow 9a^{2}-6a+1=9a^{2}-45a+27 $
Þ $ 39a=26 $
$ \Rightarrow a=\frac{2}{3} $ .
BETA
