The value of $\mathrm{k}$ for which the equation $(K-2) x^2+8 x+K+4=0$ has both roots real, distinct and…
The value of $\mathrm{k}$ for which the equation $(K-2) x^2+8 x+K+4=0$ has both roots real, distinct and negative is
6
3
4
1
Solution
$(K-2) x^2+8 x+K+4=0$
If real roots then,
$
\begin{aligned}
& 8^2-4(K-2)(K+4)>0 \\
\Rightarrow & K^2+2 K-8 < 16 \\
\Rightarrow & K^2+6 K-4 K-24 < 0 \\
\Rightarrow & (K+6)(K-4) < 0 \\
\Rightarrow & -6 < K < 4
\end{aligned}
$
If both roots are negative then $\alpha \beta$ is $+v e$
$
\begin{aligned}
& \Rightarrow \frac{K+4}{K-2}>0 \Rightarrow K>-4 \\
& \text { Also, } \frac{K-2}{K+4}>0 \Rightarrow K>2
\end{aligned}
$
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
$