Question

10．若x^a-3x^a-b+1=0是關于x的一元二次方程，求a，b的值．
下面是兩位同學的解法：
甲生：根據題意得$\left\{\begin{array}{l}{a=2}\\{a-b=1}\end{array}\right.$解方程組得$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$
乙生：依題意，得$\left\{\begin{array}{l}{a=2}\\{a-b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=1}\\{a-b=2}\end{array}\right.$，解方程組得$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$或$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$
你認為上述兩位同學的解答是否正確？為什么？如果不對，請給出正確的答案．

Answer 1

分析 本題根據一元二次方程的定義求解．分5種情況分別求解即可．

解答 解：上述兩位同學的解法都不正確，
∵x^a-3x^a-b+1=0是關于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$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$．
綜上所述，$\left\{\begin{array}{l}{a=2}\\{b=2}\end{array}\right.$；$\left\{\begin{array}{l}{a=2}\\{b=1}\end{array}\right.$；$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$；$\left\{\begin{array}{l}{a=0}\\{b=-2}\end{array}\right.$；$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$．

點評 本題主要考查了一元二次方程的概念．解題的關鍵是分5種情況討論x的指數．