關於x的一元二次方程（a﹣6）x2﹣8x+9=0有實根．                           ...

問題詳情：

關於x的一元二次方程（a﹣6）x2﹣8x+9=0有實根．                                      

（1）求a的最大整數值；                                                                       

（2）當a取最大整數值時，①求出該方程的根；②求的值．

【回答】

【考點】根的判別式；解一元二次方程-公式法．                                           

【分析】（1）根據一元二次方程的定義和根的判別式得到△=64﹣4×（a﹣6）×9≥0且a﹣6≠0，解得a≤且a≠6，然後在次範圍內找出最大的整數；                                                              

（2）①把a的值代入方程得到x2﹣8x+9=0，然後利用求根公式法求解；               

②由於x2﹣8x+9=0則x2﹣8x=﹣9，然後把x2﹣8x=﹣9整體代入所求的代數式中得到原式=2x2﹣=2x2﹣16x+，再變形得到2（x2﹣8x）+，再利用整體思想計算即可．                                  

【解答】解：（1）根據題意△=64﹣4×（a﹣6）×9≥0且a﹣6≠0，                   

解得a≤且a≠6，                                                                         

所以a的最大整數值為7；                                                                       

（2）①當a=7時，原方程變形為x2﹣8x+9=0，                                              

△=64﹣4×9=28，                                                                             

∴x=，                                                                                 

∴x1=4+，x2=4﹣；                                                                        

②∵x2﹣8x+9=0，                                                                              

∴x2﹣8x=﹣9，                                                                                 

所以原式=2x2﹣，                                                                      

=2x2﹣16x+，                                                                                  

=2（x2﹣8x）+，                                                                            

=2×（﹣9）+，                                                                              

=﹣．                                                                                           

知識點：解一元二次方程

題型：解答題