Question

The least integral value of $k$ for which $(k-2)x^2+8x+k+4>0$ for all $xϵR$, is

• A. $5$
• B. $4$
• C. $3$
• D. None of these

Answer 1

The correct option is A $5$

$(k-2).x^2+8x+(k+4)>0$

Here $b^2–4.a.c=0$.

${\left(8\right)}^2–4.(k-2).(k+4)=0$

or $(k-2)(k+4)=16$.

or $k^2+2k-8–16=0$.

or $k^2+2k-24=0$.

or $(k+6)(k-4)=0$.

$k=-6$ or $4$

$K>4$

Hence, the answer is $5$.