Complex Numbers And Quadratic Equations question 204
Question: The two roots of an equation $ x^{3}-9x^{2}+14x+24=0 $ are in the ratio 3 : 2. The roots will be [UPSEAT 1999]
Options:
A) 6, 4, - 1
B) 6, 4, 1
C) - 6, 4, 1
D) - 6, - 4, 1
Show Answer
Answer:
Correct Answer: A
Solution:
Let required roots are $ 3\alpha ,2\alpha ,\beta $ ( $ \because $ ratio of two roots are $ 3:2 $ )
$ \therefore \sum \alpha =3\alpha +2\alpha +\beta =\frac{-(-9)}{1}=9 $
Þ $ 5\alpha +\beta =9 $ ..?(i) $ \sum \alpha \beta =3\alpha .2\alpha +2\alpha .\beta +\beta .3\alpha $ $ =14 $
Þ $ 5\alpha \beta +6{{\alpha }^{2}}=14 $ …….(ii) and $ \sum \alpha \beta \gamma =3\alpha .2\alpha .\beta =-24 $
Þ $ 6{{\alpha }^{2}}\beta =-24 $ or $ {{\alpha }^{2}}\beta =-4 $ …….(iii) from (i), $ \beta =9-5\alpha , $ put the value of $ \beta $ in (ii)
Þ $ 5\alpha (9-5\alpha )+6{{\alpha }^{2}}=14 $
Þ $ 19{{\alpha }^{2}}-45\alpha +14=0 $
Þ $ (\alpha -2)(19\alpha -7)=0 $
$ \therefore $ $ \alpha =2 $ or $ \frac{7}{19} $ from (i) , if $ \alpha =2, $ then $ \beta =9-5\times 2 $ = -1 $ \because $ $ \alpha =2,\beta =-1 $ satisfy the equation (iii) so required roots are 6, 4, -1.
BETA
