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

Answer: 5

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 $5$