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

Question

If a, b ∈ R, a≠0 and the quadratic equation ax2 -bx + 1=0 has imaginary roots, then (a+b +1) is (A) positive (B) negative (C) zero (D) dependent on the sign of b

Tardigrade Admin · Accepted Answer

D=b2 -4a < 0 ⇒ a > 0 Therefore the graph is concave upwards. f(x) > 0, ∀ x∈ R ⇒ f(-1) >0 ⇒ a+b+1 >0

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

positive

negative

zero

dependent on the sign of b

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

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