Question

The graph of the function $y=16x^2+8(a+2)x-3a-2$ is strictly above the x-axis, then number of integral values of $a$ are

Answer 1

Given function $y=16x^2+8(a+2)x-3a-2$
Graph of given function lies above x-axis.
$\Rightarrow y>0$ $\forall x\in R$.
$\Rightarrow 16x^2+8(a+2)x-3a-2>0\forall x\in R$
$\Rightarrow D<0$
$\Rightarrow {\left\{8\right(a+2\left)\right\}}^2+4\times 16(3a+2)<0$
$\Rightarrow a^2+7a+6<0$
$\Rightarrow (a+6)(a+1)<0$
$\Rightarrow a\in (-6,-1)$
So, possible integral values of a are$-5,-4,-3,-2$.
$\therefore$ Number of integral values $=4$.