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  4. If f(x)=4 x2+a x+(a-3) is negative for atleast one negative x, then the possible value(s) of a, is/are

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Q. If $f(x)=4 x^2+a x+(a-3)$ is negative for atleast one negative $x$, then the possible value(s) of $a$, is/are

Complex Numbers and Quadratic Equations

Solution:

$\text { Case-I: if } f (0)<0 $
$a -3<0 \Rightarrow a <3$
$a \in(-\infty,+3) $
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Case-II: $\text { if } f (0) \geq 0 \text { and }$
$D >0$ and $-\frac{ b }{2 a }<0$
$f (0)>0$ gives $a -3 \geq 0 \Rightarrow a \geq 3$ ....(1)
image
$D>0$ gives $a^2-16(a-3)>0$
$a^2-16 a-48>0 $
$(a-12)(a-4)>0 \Rightarrow a>12 \text { or } a<4$ ....(2)
$-\frac{b}{2 a}<0$ gives $-\frac{a}{8}<0 \Rightarrow a>0$ ....(3)
$\text { from (1), (2) and (3) } $
$a \in[3,4) \cup(12, \infty)$
$\text { finally } a \in(-\infty, 4) \cup(12, \infty)$