Quadratic EquationHard
Question
If $ax^{2} + bx + 8 = 0,a,b \in R,a \neq 0$ has no distinct real roots, then the least value of $4a + b$ is
Options
A.-4
B.-3
C.-2
D.-1
Solution
$f(x) = {ax}^{2} + bx + 8 \geq 0\forall x \in R$
$$f(4) = 16a + 4\text{ }b + 8 \geq 0 \Rightarrow 4a + b \geq - 2$$
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