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

Solution

## The correct option is C $$2$$$$(K-2)x^2+8x+K+4=0$$roots are real distinct and negative$$\Rightarrow \ D\ge 0,\ -\dfrac {b}{2a} <$$ and $$af(0) > 0$$$$D=8^2-4(K-2)(K+4)$$$$=4(16-K^2-2K+4K-8)$$$$=4(16-K^2-2K+8)$$$$=4(24-2K-K^2)\ge 0$$$$\Rightarrow \ K^2+2K-24 \le 0$$$$\Rightarrow \ (12+6)(K-4)\le 0$$$$\Rightarrow \ K\in [-6,\ 4]----(1)$$$$\dfrac -{b}{2a}=\dfrac {-8}{2(K-2)} < 0$$$$\Rightarrow \ \dfrac {1}{K-2} > 0\ \Rightarrow \ K > 2-----(2)$$$$af(0)=(K-2)(K+4) >$$$$\Rightarrow \ \in (-\infty ,\ -4)\cap (2,\ \infty)-----(3)$$From $$(1),\ (2)$$ and $$(3)$$$$K \in (2,\ 4)$$$$\Rightarrow \ K=3,\ 4$$ are integer.Maths

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