Complex Numbers And Quadratic Equations question 602
Question: Let $ \alpha ,\beta $ be the roots of $ x^{2}-x+p=0 $ and $ \gamma ,\delta $ be the roots of $ x^{2}-4x+q=0 $ . If $ \alpha ,\beta ,\gamma ,\delta $ are in G.P., then integral values of $ p,q $ are respectively [IIT Screening 2001]
Options:
A) - 2, - 32
B) - 2, 3
C) - 6, 3
D) - 6, - 32
Show Answer
Answer:
Correct Answer: A
Solution:
Let r be the common ratio of the G.P. a, b, g, d then $ \beta =\alpha r, $ $ \gamma =\alpha r^{2} $ and $ \delta =\alpha r^{3} $
$ \therefore \alpha +\beta =1 $
$ \Rightarrow \alpha +\alpha r=1 $
$ \Rightarrow \alpha (1+r)=1 $ …….(i) $ \alpha \beta =p\Rightarrow \alpha (\alpha r)=p\Rightarrow {{\alpha }^{2}}r=p $ …….(ii) $ \gamma +\delta =4\Rightarrow \alpha r^{2}+\alpha r^{3}=4 $ $ \gamma +\delta =q\Rightarrow ar^{2}ar^{3} $ $ \alpha r^{2}(1+r)=4 $ …….(iii) and $ \gamma \delta =q\Rightarrow \alpha r^{2}.\alpha r^{3}=q $
$ \Rightarrow {{\alpha }^{2}}r^{5}=q $ …….(iv) Dividing (iii) by (i), we get, $ r^{2}=4\Rightarrow r=\pm 2 $ If we take $ r=2 $ , then $ \alpha $ is not integral, so we take $ r=-2, $ Substituting $ r=-2 $ in (i), we get $ \alpha =-1 $ Now, from (ii), we have $ p={{\alpha }^{2}}r={{(-1)}^{2}}(-2)=-2 $ and from (iv), we have $ q={{\alpha }^{2}}r^{5}={{(-1)}^{2}}{{(-2)}^{5}}=-32 $
Þ (p, q) = (- 2, - 32).
BETA
