Question

4．若x^2a+b-2x^a-b+3=0是关于x的一元二次方程，求a，b的值．

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：∵x^2a+b-2x^a+b+3=0是关于x的一元二次方程，
∴①$\left\{\begin{array}{l}{2a+b=2}\\{a-b=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=1}\\{b=0}\end{array}\right.$；
②$\left\{\begin{array}{l}{2a+b=2}\\{a-b=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{2}{3}}\\{b=\frac{2}{3}}\end{array}\right.$；
③$\left\{\begin{array}{l}{2a+b=1}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$；
④$\left\{\begin{array}{l}{2a+b=0}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{2}{3}}\\{b=-\frac{4}{3}}\end{array}\right.$；
⑤$\left\{\begin{array}{l}{2a+b=2}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{4}{3}}\\{b=-\frac{2}{3}}\end{array}\right.$；
综上所述$\left\{\begin{array}{l}{a=1}\\{b=0}\end{array}\right.$，$\left\{\begin{array}{l}{a=\frac{2}{3}}\\{b=\frac{2}{3}}\end{array}\right.$，$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$，$\left\{\begin{array}{l}{a=\frac{2}{3}}\\{b=-\frac{4}{3}}\end{array}\right.$，$\left\{\begin{array}{l}{a=\frac{4}{3}}\\{b=-\frac{2}{3}}\end{array}\right.$．

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．