Mathematics - Quadratic Equation Question with Solution | TestHub

For $x\ge 4$, the last digit of $1!+2!+\dots +x!$ is 3.

For $x<4$, the given equation has only solutions:

$$\begin{array}{rl}& x=1,K=\pm 1\text{ and }x=3,K=\pm 3\\ & \alpha =1,\beta =3\\ & (a+\sqrt{b}{)}^{x^2-15}+(a-\sqrt{b}{)}^{x^2-15}=2a\\ & \therefore x^2-15=\pm 1\Rightarrow x=\pm 4,\pm \sqrt{14}\\ & {\alpha }_1=-4,{\alpha }_2=4,{\alpha }_3=-\sqrt{14},{\alpha }_4=\sqrt{14}\\ & \therefore \left|{\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4-{\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4\right|=|0-16\times 14|=224\end{array}$$