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)

$3$

Answer 2

$(K-2) x^2 + 8x + K + 4 = 0$ 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 $K$. $\therefore K=3$