Mathematics - Quadratic Equation Question with Solution | TestHub

The discriminant of the quadratic equation is $D=(-(2k-1){)}^2-4(1)(k^2-k-6)=4k^2-4k+1-4k^2+4k+24=25$. Since $D=25>0$, the roots are always real and distinct. The roots are $\alpha ,\beta =\frac{(2k-1)\pm \sqrt{25}}{2}=\frac{2k-1\pm 5}{2}$. Thus, the roots are $\alpha =k+2$ and $\beta =k-3$. A. For both roots to be positive, $k+2>0⟹k>-2$ and $k-3>0⟹k>3$. Both conditions imply $k>3$. So, $k\in (3,\infty )$. This statement is correct. B. For both roots to be negative, $k+2<0⟹k<-2$ and $k-3<0⟹k<3$. Both conditions imply $k<-2$. So, $k\in (-\infty ,-2)$. This statement is correct. C. For exactly one root to lie in $(-1,1)$, we have two cases: Case 1: $k+2\in (-1,1)$ and $k-3\notin (-1,1)$. $-1<k+2<1⟹-3<k<-1$. For these values, $k-3\in (-6,-4)$, which is not in $(-1,1)$. So $k\in (-3,-1)$ is a valid range. Case 2: $k-3\in (-1,1)$ and $k+2\notin (-1,1)$. $-1<k-3<1⟹2<k<4$. For these values, $k+2\in (4,6)$, which is not in $(-1,1)$. So $k\in (2,4)$ is a valid range. Combining both cases, $k\in (-3,-1)\cup (2,4)$. This statement is correct. D. The absolute difference between the roots is $|\alpha -\beta |=|(k+2)-(k-3)|=|5|=5$. Since $5\ngtr 6$, this statement is incorrect.