Mathematics - Quadratic Equation Question with Solution | TestHub

For the quadratic equation to have integer roots, its discriminant must be a perfect square, say $k^2\ge 0$. The discriminant is $D=(m+1{)}^2-4(m^2+m-8)=m^2+2m+1-4m^2-4m+32=-3m^2-2m+33$. So, $-3m^2-2m+33=k^2$. For $D\ge 0$, we must have $3m^2+2m-33\le 0$. The roots of $3m^2+2m-33=0$ are $m=\frac{-2\pm \sqrt{4-4(3)(-33)}}{6}=\frac{-2\pm 20}{6}$, which are $m=3$ and $m=-11/3$. Thus, $m\in [-11/3,3]$. The integer values of $m$ in this interval are $\{-3,-2,-1,0,1,2,3\}$. We test these values for $D=k^2$: For $m=-3,D=12$. For $m=-2,D=25=5^2$. This is a valid $m$. Roots are $x=\frac{(-2+1)\pm 5}{2}=\frac{-1\pm 5}{2}$, which are $2,-3$ (integers). For $m=-1,D=32$. For $m=0,D=33$. For $m=1,D=28$. For $m=2,D=17$. For $m=3,D=0=0^2$. This is a valid $m$. Roots are $x=\frac{(3+1)\pm 0}{2}=\frac{4}{2}$, which are $2,2$ (integers). Only $m=-2$ and $m=3$ yield integer roots. Therefore, there are 2 such integer values of $m$.