Question

10．若x^a-3x^a-b+1=0是关于x的一元二次方程，求a，b的值．
下面是两位同学的解法：
甲生：根据题意得$\left\{\begin{array}{l}{a=2}\\{a-b=1}\end{array}\right.$解方程组得$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$
乙生：依题意，得$\left\{\begin{array}{l}{a=2}\\{a-b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=1}\\{a-b=2}\end{array}\right.$，解方程组得$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$
你认为上述两位同学的解答是否正确？为什么？如果不对，请给出正确的答案．

Answer 1

分析 本题根据一元二次方程的定义求解．分5种情况分别求解即可．

解答 解：上述两位同学的解法都不正确，
∵x^a-3x^a-b+1=0是关于x的一元二次方程，
∴①$\left\{\begin{array}{l}{a=2}\\{a-b=0}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=2}\\{b=2}\end{array}\right.$；
②$\left\{\begin{array}{l}{a=2}\\{a-b=1}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$；
③$\left\{\begin{array}{l}{a=2}\\{a-b=2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$；
④$\left\{\begin{array}{l}{a=0}\\{a-b=2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=0}\\{b=-2}\end{array}\right.$；
⑤$\left\{\begin{array}{l}{a=1}\\{a-b=2}\end{array}\right.$，
解得$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$．
综上所述，$\left\{\begin{array}{l}{a=2}\\{b=2}\end{array}\right.$；$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$；$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$；$\left\{\begin{array}{l}{a=0}\\{b=-2}\end{array}\right.$；$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$．

点评 本题主要考查了一元二次方程的概念．解题的关键是分5种情况讨论x的指数．