First slide

Theory of equations

Question

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

Easy

A

positive
B

negative
C

zero
D

dependent on the sign of b

Solution

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

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