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  4. If a, b ∈ R, a≠0 and the quadratic equation ax2 -bx + 1=0 has imaginary roots, then (a+b +1) is

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Q. If $a, b \in R, a\ne0$ and the quadratic equation $ax^{2} -bx + 1=0$ has imaginary roots, then $(a+b +1)$ is

Complex Numbers and Quadratic Equations

Solution:

$D=b^{2} -4a < 0 \Rightarrow a > 0$
Therefore the graph is concave upwards.
$f\left(x\right) > 0, \forall x\in R$
$\Rightarrow f\left(-1\right) >0 $
$\Rightarrow a+b+1 >0$