Complex Numbers And Quadratic Equations question 800
Question: The values of ‘a’ and ‘b’ for which equation $ x^{4}-4x^{3}+ax^{2}+bx+1=0 $ have four real roots [DCE 2005]
Options:
A) - 6, - 4
B) - 6, 5
C) - 6, 4
D) 6, - 4
Show Answer
Answer:
Correct Answer: D
Solution:
Let for real roots are $ \alpha ,\beta ,\gamma ,\delta $ then equation is $ (x-\alpha )(x-\beta )(x-\gamma )(x-\delta )=0 $ $ x^{4}-(\alpha +\beta +\gamma +\delta )x^{3}+(\alpha \beta +\beta \gamma +\gamma \delta $ $ +\alpha \delta +\beta \delta +\alpha \gamma )x^{2}-(\alpha \beta \gamma +\beta \gamma \delta $ $ +\alpha \beta \delta +\alpha \gamma \delta )x+\alpha \beta \gamma \delta =0 $ $ x^{4}-\sum \alpha .x^{3}+\sum \alpha \beta .x^{2}-\sum \alpha \beta \gamma .x+\alpha \beta \gamma \delta =0 $ on comparing with $ x^{4}-4x^{3}+ax^{2}+bx+1=0 $ $ \sum \alpha =4,\sum \alpha \beta =a $ $ \sum \alpha \beta \gamma =-b,\alpha \beta \gamma \delta =1 $ For real roots, A.M. of roots $ \ge $ G.M. of roots $ \frac{1}{4}(\sum \alpha )\ge {{(\alpha \beta \gamma \delta )}^{1/4}} $ ; $ \sum \alpha =4 $
$ \therefore $ $ \frac{1}{4}\sum \alpha =\frac{1}{4}\times 4=1 $ $ {{(\alpha \beta \gamma \delta )}^{1/4}}=1 $
Þ $ \alpha \beta \gamma \delta =1 $ $ \Sigma \alpha =4 $ and $ \alpha \beta \gamma \delta =1 $
$ \therefore $ $ \alpha =\beta =\gamma =\delta $ = 1 Now, $ \sum \alpha \beta =a $
$ \therefore a=\alpha \beta +\beta \gamma +\gamma \delta +\alpha \delta +\beta \delta +\alpha \gamma $ $ =1\times 1+1\times 1+1\times 1+1\times 1+1\times 1+1\times 1 $ = 6 $ -b=\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta $ $ =1\times 1\times 1+1\times 1\times 1+1\times 1\times 1+1\times 1\times 1 $ $ ={{(1)}^{3}}+{{(1)}^{3}}+{{(1)}^{3}}+{{(1)}^{3}}=1+1+1+1=4 $
$ \therefore b=-4 $ ;
$ \therefore a=6 $ and $ b=-4 $ .
BETA
