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 $K$.
$\therefore K=3$