Complex Numbers And Quadratic Equations question 32
Question: If $ b>a $ , then the equation $ (x-a)(x-b)=1 $ has [IIT Screening 2000]
Options:
A) Both roots in $ [a,b] $
B) Both roots in $ (-\infty ,a) $
C) Both roots in $ (b,+\infty ) $
D) One root in $ (-\infty ,a) $ and the other in $ (b,+\infty ) $
Show Answer
Answer:
Correct Answer: D
Solution:
The equation is $ x^{2}-(a+b)x+ab-1=0 $
$ \therefore $ discriminant $ ={{(a+b)}^{2}}-4(ab-1)={{(b-a)}^{2}}+4>0 $
$ \therefore $ both roots are real. Let them be $ \alpha ,\beta $ where $ \alpha =\frac{(a+b)-\sqrt{{{(b-a)}^{2}}+4}}{2} $ , $ \beta =\frac{(a+b)+\sqrt{{{(b-a)}^{2}}+4}}{2} $ Clearly, $ \alpha <\frac{(a+b)-\sqrt{{{(b-a)}^{2}}}}{2}=\frac{(a+b)-(b-a)}{2}=a $ $ (\because b>a) $ and $ \beta >\frac{(a+b)+\sqrt{{{(b-a)}^{2}}}}{2}=\frac{a+b+b-a}{2}=b $ Hence, one root $ \alpha $ is less than a and the other root b is greater than b.
BETA
