关于x的一元二次方程（a﹣6）x2﹣8x+9=0有实根．                           ...

问题详情：

关于x的一元二次方程（a﹣6）x2﹣8x+9=0有实根．                                      

（1）求a的最大整数值；                                                                       

（2）当a取最大整数值时，①求出该方程的根；②求的值．

【回答】

【考点】根的判别式；解一元二次方程-公式法．                                           

【分析】（1）根据一元二次方程的定义和根的判别式得到△=64﹣4×（a﹣6）×9≥0且a﹣6≠0，解得a≤且a≠6，然后在次范围内找出最大的整数；                                                              

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

②由于x2﹣8x+9=0则x2﹣8x=﹣9，然后把x2﹣8x=﹣9整体代入所求的代数式中得到原式=2x2﹣=2x2﹣16x+，再变形得到2（x2﹣8x）+，再利用整体思想计算即可．                                  

【解答】解：（1）根据题意△=64﹣4×（a﹣6）×9≥0且a﹣6≠0，                   

解得a≤且a≠6，                                                                         

所以a的最大整数值为7；                                                                       

（2）①当a=7时，原方程变形为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）+，                                                                              

=﹣．                                                                                           

知识点：解一元二次方程

题型：解答题