Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

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

roots are real distinct and negative

$\Rightarrow D\ge 0,-\frac{b}{2a}<$ and $af\left(0\right)>0$

$D=8^2-4(K-2)(K+4)$

$=4(16-K^2-2K+4K-8)$

$=4(16-K^2-2K+8)$

$=4(24-2K-K^2)\ge 0$

$\Rightarrow K^2+2K-24\le 0$

$\Rightarrow (12+6)(K-4)\le 0$

$\Rightarrow K\in [-6,4]----\left(1\right)$

$\frac{-}{b}2a=\frac{-8}{2(K-2)}<0$

$\Rightarrow \frac{1}{K-2}>0\Rightarrow K>2-----\left(2\right)$

$af\left(0\right)=(K-2)(K+4)>$

$\Rightarrow \in (-\infty ,-4)\cap (2,\infty )-----\left(3\right)$

From $\left(1\right),\left(2\right)$ and $\left(3\right)$

$K\in (2,4)$

$\Rightarrow K=3,4$ are integer.