Mathematics - Quadratic Equation Question with Solution | TestHub

For a quadratic equation $A{x}^2+Bx+C=0$ to have no real roots, two conditions must be met: 1) The coefficient of $x^2$ must not be zero, so $p-1\ne 0⟹p\ne 1$. 2) The discriminant $D=B^2-4AC$ must be less than 0. Here, $A=(p-1)$, $B=(p-1)$, and $C=1$. So, $D=(p-1{)}^2-4(p-1)(1)<0$. Factoring out $(p-1)$, we get $(p-1)[(p-1)-4]<0$, which simplifies to $(p-1)(p-5)<0$. This inequality holds when $1<p<5$. Combining this with the condition $p\ne 1$, the set of all possible values of $p$ is $(1,5)$.