Question

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

• A. $6$
• B. $3$
• C. $4$
• D. $1$

Answer 1

Answer: (B)

$(K-2)x^2+8x+K+4=0$
If real roots then,
$8^2$
$-4\left(K-2\right)\left(K+4\right)>0$
$\Rightarrow K^2+2K-S<16$
$\Rightarrow \left(K+6\right)\left(K-4\right)<0$
$\Rightarrow -6<K<4$
If both roots are negative then $\alpha \beta$ is $+ve$
$\Rightarrow \frac{K+4}{K-2}>0\Rightarrow K>-4$
Also, $\frac{K-2}{K+4}>0\Rightarrow K>2$
Roots are real so, $-6<K<4$
So, 6 and 4 are not correct.
Since, $K>2$, so 1 is also not correct value of $K$.
$\therefore K=3$