Mathematics - Quadratic Equation Question with Solution | TestHub

For the roots of a quadratic equation to be real and distinct, its discriminant $D$ must be strictly greater than zero. For the given equation $x^2-2(m+1)x+m^2+4=0$, the discriminant is $D=(-2(m+1){)}^2-4(1)(m^2+4)$. We set $D>0$: $4(m+1{)}^2-4(m^2+4)>0$. Dividing by 4, we get $(m+1{)}^2-(m^2+4)>0$. Expanding this, $m^2+2m+1-m^2-4>0$. This simplifies to $2m-3>0$. Solving for $m$, we get $2m>3$, or $m>3/2$. Therefore, the range of $m$ is $(3/2,\infty )$.