Mathematics - Quadratic Equation Question with Solution | TestHub

$(K-2)x^2+8x+K+4=0$ If real roots then, $$\begin{array}{rl}& 8^2-4(K-2)(K+4)>0\\ \Rightarrow & K^2+2K-8<16\\ \Rightarrow & K^2+6K-4K-24<0\\ \Rightarrow & (K+6)(K-4)<0\\ \Rightarrow & -6<K<4\end{array}$$ If both roots are negative then $\alpha \beta$ is $+ve$ $$\begin{array}{rl}& \Rightarrow \frac{K+4}{K-2}>0\Rightarrow K>-4\\ & \text{ Also, }\frac{K-2}{K+4}>0\Rightarrow K>2\end{array}$$ 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 $K$. $$\therefore K=3$$