Mathematics - Quadratic Equation Question with Solution | TestHub

For the roots of $a{x}^2+bx+c=0$ to be integers, the discriminant $D=b^2-4ac$ must be a perfect square, say $N^2$, where $N\ge 0$. Here, $D=(k+5{)}^2-4(k^2-5k+6)=k^2+10k+25-4k^2+20k-24=-3k^2+30k+1$. We need $-3k^2+30k+1=N^2$. Since $N^2\ge 0$, we must have $-3k^2+30k+1\ge 0$, which implies $3k^2-30k-1\le 0$. The roots of $3k^2-30k-1=0$ are $k=\frac{30\pm \sqrt{900+12}}{6}=\frac{30\pm \sqrt{912}}{6}$. Since $\sqrt{900}=30$ and $\sqrt{961}=31$, $\sqrt{912}\approx 30.19$. So $k\in [\frac{30-30.19}{6},\frac{30+30.19}{6}]\approx [-0.03,10.03]$. Integer values for $k$ are $0,1,...,10$. We check which of these make $D$ a perfect square: $k=0⟹D=1=1^2$. $k=2⟹D=-3(4)+30(2)+1=49=7^2$. $k=3⟹D=-3(9)+30(3)+1=64=8^2$. $k=7⟹D=-3(49)+30(7)+1=64=8^2$. $k=8⟹D=-3(64)+30(8)+1=49=7^2$. $k=10⟹D=-3(100)+30(10)+1=1=1^2$. For other integer values of $k$ in the range, $D$ is not a perfect square. The possible integer values of $k$ are $0,2,3,7,8,10$. The sum of these values is $0+2+3+7+8+10=30$.