Mathematics - Quadratic Equation Question with Solution | TestHub

For real and distinct roots, the discriminant $D>0$ and the coefficient of $x^2$ must not be zero. $D=(-(k+2){)}^2-4(k-1)(k)=(k+2{)}^2-4k^2+4k=k^2+4k+4-4k^2+4k=-3k^2+8k+4$. We need $-3k^2+8k+4>0$, which implies $3k^2-8k-4<0$. The roots of $3k^2-8k-4=0$ are $k=\frac{8\pm \sqrt{64-4(3)(-4)}}{6}=\frac{8\pm \sqrt{112}}{6}=\frac{8\pm 4\sqrt{7}}{6}=\frac{4\pm 2\sqrt{7}}{3}$. Thus, $k\in \left(\frac{4-2\sqrt{7}}{3},\frac{4+2\sqrt{7}}{3}\right)$. Also, for the equation to be quadratic, $k-1\ne 0$, so $k\ne 1$. Since $1$ lies within the interval $\left(\frac{4-2\sqrt{7}}{3},\frac{4+2\sqrt{7}}{3}\right)$ (approximately $(-0.43,3.09)$), we must exclude $k=1$. Hence, $k\in \left(\frac{4-2\sqrt{7}}{3},\frac{4+2\sqrt{7}}{3}\right)\setminus \{1\}$.