Complex Numbers And Quadratic Equations question 199
Question: The values of a for which $ 2x^{2}-2(2a+1)x+a(a+1)=0 $ may have one root less than a and other root greater than a are given by [UPSEAT 2001]
Options:
A) $ 1>a>0 $
B) $ -1<a<0 $
C) $ a\ge 0 $
D) $ a>0\text{or }a<-1 $
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Answer:
Correct Answer: D
Solution:
The given condition suggest that a lies between the roots. Let $ f(x)=2x^{2}-2(2a+1)x+a(a+1) $ For ?a? to lie between the roots we must have Discriminant $ \ge 0 $ and $ f(a)<0 $ . Now, Discriminant $ \ge 0 $
$ \Rightarrow 4{{(2a+1)}^{2}}-8a(a+1)\ge 0 $
$ \Rightarrow 8(a^{2}+a+1/2)\ge 0 $ which is always true. Also $ f(a)<0\Rightarrow 2a^{2}-2a(2a+1)+a(a+1)<0 $
$ \Rightarrow -a^{2}-a<0 $
$ \Rightarrow a^{2}+a>0\Rightarrow a(1+a)>0 $
$ \Rightarrow a>0 $ or $ a<-1 $ .
BETA
