Mathematics - Quadratic Equation Question with Solution | TestHub

For the quadratic expression $A{x}^2+Bx+C$ to be positive for all real $x$, two conditions must be met: 1. The leading coefficient $A$ must be positive. 2. The discriminant $D=B^2-4AC$ must be negative. In this case, $A=k^2-1$, $B=k-1$, $C=1$. Condition 1: $k^2-1>0⟹(k-1)(k+1)>0$. This implies $k\in (-\infty ,-1)\cup (1,\infty )$. Condition 2: $D=(k-1{)}^2-4(k^2-1)(1)<0$. Factor out $(k-1)$: $(k-1)[(k-1)-4(k+1)]<0⟹(k-1)[-3k-5]<0$. Multiply by $-1$ and reverse the inequality sign: $(k-1)(3k+5)>0$. This implies $k\in (-\infty ,-5/3)\cup (1,\infty )$. To satisfy both conditions, we need to find the intersection of the two intervals: $((-\infty ,-1)\cup (1,\infty ))\cap ((-\infty ,-5/3)\cup (1,\infty ))$. Since $-5/3\approx -1.67$, the intersection is $k\in (-\infty ,-5/3)\cup (1,\infty )$. We are looking for integer values of $k$ in the interval $[-10,10]$. For $k\in (-\infty ,-5/3)$, the integers in $[-10,10]$ are $\{-10,-9,-8,-7,-6,-5,-4,-3,-2\}$. There are $9$ such integers. For $k\in (1,\infty )$, the integers in $[-10,10]$ are $\{2,3,4,5,6,7,8,9,10\}$. There are $9$ such integers. The total number of integer values of $k$ is $9+9=18$.