Question

Find the least integral value of $k$ for which, $\left(k - 2\right)x^{2}+8x+k+4>0$ for all $x\in R.$

Answer 1

We have, $\left(k - 2\right)x^{2}+8x+k+4>0$
Now, $a>0$ and $D < 0$
$\Rightarrow k-2>0$ and $64-4\left(k - 2\right)\left(k + 4\right) < 0$
$\Rightarrow k>2$ and $k^{2}+2k-24>0$
$\Rightarrow k>2$ and $\left(k + 6\right)\left(k - 4\right)>0$

Therefore, $k\in \left(4 , \infty \right)$
Hence, least integral value of $k$ is $5$