Question

The least integral value of $k$ for which $(k-1)x^2+8x+k+5$ is always positive $\forall x\in R,$ is

Answer 1

$(k-1)x^2+8x+k+5>0\forall x\in R,$
When $k=1$, we get
$8x+6>0\Rightarrow x>-\frac{3}{4}$
Which is not valid for all $x\in R$

Now,
$k-1>0\text{and}D<0$
Now,
$k-1>0\Rightarrow k>1\cdots \left(1\right)$

$D<0$
$\Rightarrow 64-4(k-1)(k+5)<0\Rightarrow 16-(k^2+4k-5)<0\Rightarrow -k^2-4k+21<0\Rightarrow k^2+4k-21>0\Rightarrow (k+7)(k-3)>0\Rightarrow k\in (-\infty ,-7)\cup (3,\infty )\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$k\in (3,\infty )$
Hence, the least integral value of $k$ is $4$.