The least integral value of k for which $(k-2) x^2 + 8x + k +4≥0$ for all x ∈ R, is _____.
Answer & explanation
We have,
$(k-2) x^2 + 8x + k +4≥0$ for all x ∈ R
$⇒k-2>0$ and $64-4 (k-2) (k + 4) ≤0$
$⇒k>2$ and $k^2 + 2k-24 ≥0$
$⇒k>2$ and $(k + 6) (k −4) ≥0$
$⇒k>2$ and $k ≤-6$ or, $k ≥4⇒ k ≥4$
Hence, the least integral value of k is 4.