Mathematics - Quadratic Equation Question with Solution | TestHub

For the equation $x^2-2(k+1)x+(k^2+2k-1)=0$, the discriminant is $D=(2(k+1){)}^2-4(k^2+2k-1)=4(k^2+2k+1)-4(k^2+2k-1)=4(2)=8$. Since $D=8>0$, the roots are always distinct real roots for any real $k$. The sum of roots $\alpha +\beta =2(k+1)$ and product of roots $\alpha \beta =k^2+2k-1$. We have $m={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(2(k+1){)}^2-2(k^2+2k-1)=4(k^2+2k+1)-2k^2-4k+2=2k^2+4k+6$. We are given that $m$ is a perfect square less than or equal to 100. So $2k^2+4k+6=N^2$ for some integer $N$ such that $N^2\le 100$. Rearranging, $2(k^2+2k+3)=N^2⟹2((k+1{)}^2+2)=N^2$. This implies $N^2$ must be an even number, so $N$ must be even. Possible values for $N^2$ are $4,16,36,64$. If $N^2=4$, $2((k+1{)}^2+2)=4⟹(k+1{)}^2+2=2⟹(k+1{)}^2=0⟹k=-1$. If $N^2=16$, $2((k+1{)}^2+2)=16⟹(k+1{)}^2+2=8⟹(k+1{)}^2=6$, no integer $k$. If $N^2=36$, $2((k+1{)}^2+2)=36⟹(k+1{)}^2+2=18⟹(k+1{)}^2=16⟹k+1=\pm 4⟹k=3$ or $k=-5$. If $N^2=64$, $2((k+1{)}^2+2)=64⟹(k+1{)}^2+2=32⟹(k+1{)}^2=30$, no integer $k$. The possible integer values of $k$ are $-1,3,-5$. There are 3 such values.