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If a ,b in R ,a!=0 and the quadratic eq...

If `a ,b in R ,a!=0` and the quadratic equation `a x^2-b x+1=0` has imaginary roots, then `(a+b+1)` is a. positive b. negative c. zero d. Dependent on the sign of `b`

A

positive

B

negative

C

zero

D

dependent on the sign of b

Text Solution

Verified by Experts

The correct Answer is:
1
`D = b^(2) - 4a lt 0 rArr a gt 0`
Therefore the graph is concave upwards.
`f(x) gt 0, AA x in R`
`rArr f(-1) gt 0`
`rArr a + b + 1 gt 0`
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