Question

If $(k-2)x^2+8x+k+4=0$ has both roots real, distinct and negative, then k =

• A. $6$
• B. $4$
• C. $3$
• D. $1$

Answer 1

Answer: (C), (D)

$3$
$1$

For roots to be real, distinct and possibly negative, $\Delta >0$

$\Delta =b^2-4ac$

$=(8{)}^2-4(k-2\left)\right(k+4)$

$=64-4(k^2+2k-8)$

$=64-4k^2-8k+32$

$=96-4k^2-8k$

Since, $\Delta >0$

$\Rightarrow$ $96-4k^2-8k>0$

$\Rightarrow$ $4k^2+8k-96<0$

$\Rightarrow$ $(4k+24)(k-4)<0$

$\Rightarrow$ $4(k+6)(k-4)<0$

$\Rightarrow$ $(k+6)(k-4)<0$

$\therefore$ Only integers between $-6<k<4$ can the roots be negative, distinct and real.