Complex Numbers And Quadratic Equations question 466
Question: If a, b $ \in $ R, $ a\ne 0 $ and the quadratic equation $ ax^{2}-bx+1=0 $ has imaginary roots then $ (a+b+1) $ is
Options:
A) positive
B) negative
C) zero
D) dependent on the sign of b
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ D=b^{2}-4a<0\Rightarrow a>0 $ Therefore the graph is concave upwards. $ f(x)>0,\forall x\in R $
$ \Rightarrow f(-1)>0 $
$ \Rightarrow a+b+1>0 $
BETA
