Question

2．已知方程2x^a+b-x^a-b-ab=0是關(guān)于x的一元二次方程，求a、b的值．

Answer 1

分析 本題根據(jù)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：∵2x^a+b-x^a-b+ab=0是關(guān)于x的一元二次方程，
∴①$\left\{\begin{array}{l}{a+b=2}\\{a-b=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=\frac{1}{2}}\end{array}\right.$；
②$\left\{\begin{array}{l}{a+b=2}\\{a-b=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=1}\\{b=1}\end{array}\right.$；
③$\left\{\begin{array}{l}{a+b=1}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=-\frac{1}{2}}\end{array}\right.$；
④$\left\{\begin{array}{l}{a+b=0}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$；
⑤$\left\{\begin{array}{l}{a+b=2}\\{a-b=2}\end{array}\right.$，解得$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$；
綜上所述：a、b的值為：$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=\frac{1}{2}}\end{array}\right.$；$\left\{\begin{array}{l}{a=1}\\{b=1}\end{array}\right.$；$\left\{\begin{array}{l}{a=\frac{3}{2}}\\{b=-\frac{1}{2}}\end{array}\right.$；$\left\{\begin{array}{l}{a=1}\\{b=-1}\end{array}\right.$；$\left\{\begin{array}{l}{a=2}\\{b=0}\end{array}\right.$．

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