Mathematics - Quadratic Equation Question with Solution | TestHub

From Vieta's formulas, $\alpha +\beta =p+2$ and $\alpha \beta =p^2-2p+1=(p-1{)}^2$. For real and distinct roots, the discriminant $D>0$. So, $D=(p+2{)}^2-4(p-1{)}^2>0$. This simplifies to $(p+2-2(p-1))(p+2+2(p-1))>0$, which gives $(-p+4)(3p)>0$. Thus, $3p(4-p)>0$, implying $p(p-4)<0$. Therefore, $0<p<4$. The possible integer values for $p$ are $1,2,3$. Next, we are given that ${\alpha }^2+{\beta }^2$ is an integer. We calculate ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(p+2{)}^2-2(p-1{)}^2$. Expanding this, we get $(p^2+4p+4)-2(p^2-2p+1)=p^2+4p+4-2p^2+4p-2=-p^2+8p+2$. Since $p$ is an integer, the expression $-p^2+8p+2$ will always be an integer. Therefore, all integer values of $p$ satisfying $0<p<4$ are valid. The possible integer values of $p$ are $1,2,3$. The sum of all possible integer values of $p$ is $1+2+3=6$.