Mathematics - Quadratic Equation Question with Solution | TestHub

The discriminant of $P(x)$ is $D=(-(2m+1){)}^2-4(1)(m^2+m-2)=4m^2+4m+1-4m^2-4m+8=9$. Since $D=9>0$ for all real $m$, $P(x)$ always has two distinct real roots. So, A is correct. The minimum value of $P(x)$ is $-D/(4a)=-9/(4\cdot 1)=-9/4$. Since $-9/4\le 0$, B is correct. The roots are $x=\frac{(2m+1)\pm \sqrt{9}}{2}=\frac{2m+1\pm 3}{2}$. So, the roots are $x_1=\frac{2m+1-3}{2}=m-1$ and $x_2=\frac{2m+1+3}{2}=m+2$. For both roots to be greater than $-1$, we need $m-1>-1⟹m>0$ and $m+2>-1⟹m>-3$. Both conditions imply $m>0$. The statement says $m>1$, which is not equivalent to $m>0$. So, C is incorrect. For one root to be positive and the other negative, the product of roots must be negative: $(m-1)(m+2)<0$. This implies $-2<m<1$. So, D is correct.