Question

The integral values of $k$ for which the equation $(k-2)x^2+8x+k+4=0$ has both the roots real, distinct and negative is:

Answer 1

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

for roots to be real, distinct and possibly negative

$\Delta >0$

$\Delta =b^2-4ac$

$\Delta =8^2-4(k-2)(k+4)$

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

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

since $\Delta >0$

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

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

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

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

Therefore, only integers between $-6<k<4$ can the roots to be negative, distinct and real.