CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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
loader
B
1
loader
C
2
loader
D
3
loader

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image