21.2.5一元二次方程的根与系数的关系 课件 -2026-2027学年华东师大版数学九年级上册

#华东师大版九年级上册数学《一元二次方程的根与系数的关系》专项练习（全文1444字）##一\(x_1、x_2\)知识点梳理对于一元二次方程$\boldsymbol{ax^2+bx+c=0(a\neq0)}$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)当$\Delta=b^2-4ac\ge0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)方程的两根为$x_1\(x_1、x_2\)x_2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)根与系数的关系（韦达定理）：$$\boldsymbol{x_1+x_2=-\dfrac{b}{a}},\quad\boldsymbol{x_1x_2=\dfrac{c}{a}}$$特殊形式：若方程为$x^2+px+q=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$x_1+x_2=-p\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)x_1x_2=q$。常用代数式变形（考试高频）：1.$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$2.$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2$3.$\dfrac1{x_1}+\dfrac1{x_2}=\dfrac{x_1+x_2}{x_1x_2}$重要前提：**使用韦达定理\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)必须同时满足两个条件：①$a≠0$（是一元二次方程）；②$\Delta\ge0$（方程存在实数根）**。很多题目只给两根关系求参数\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)最后一定要检验判别式。韦达定理常见题型：已知方程\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求两根和\(x_1、x_2\)两根积；已知一根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求另一根与参数；不解方程\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求含两根的代数式的值；已知两根构造一元二次方程。构造方程：以$x_1,x_2$为两根的一元二次方程为$x^2-(x_1+x_2)x+x_1x_2=0$。本节高频易错点：①忽略$a≠0$或者不检验$\Delta\ge0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求出参数后直接写答案；②记反公式\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)把$x_1+x_2$写成$\dfrac{b}{a}$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)漏掉负号；③代数式变形出错\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1^2+x_2^2$忘记减$2x_1x_2$；②代入系数时符号出错；④题目没有说明有实数根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)直接套用韦达定理；⑤两根异号判断：$x_1x_2<0$时\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$\Delta$一定大于0\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)不用额外检验判别式。##二\(x_1、x_2\)基础选择题1.一元二次方程$x^2-4x+3=0$两根$x_1,x_2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$x_1+x_2$的值为（）A.$4$B.$-4$C.$3$D.$-3$2.方程$2x^2+5x-3=0$的两根之积$x_1x_2$是（）A.$\dfrac52$B.$-\dfrac52$C.$\dfrac32$D.$-\dfrac32$3.已知方程$x^2+kx-6=0$一个根是$2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则另一根为（）A.$3$B.$-3$C.$6$D.$-6$4.若一元二次方程$x^2-2x+\(m=\)0$两根满足$x_1+x_2=2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1x_2=m$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)两根平方和为6\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$\(m=\)$（）A.$-1$B.$1$C.$-2$D.$2$5.已知方程$x^2-3x+c=0$两根互为倒数\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$c$的值为（）A.$0$B.$1$C.$2$D.$3$##三\(x_1、x_2\)填空题1.方程$x^2+6x-2=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1+x_2=$____\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1x_2=$____。2.一元二次方程$3x^2-2x-5=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$\dfrac1{x_1}+\dfrac1{x_2}=$________。3.已知$x_1,x_2$是方程$x^2-5x+1=0$两根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$x_1^2+x_2^2=$________。4.以$2$和$-3$为两根的一元二次方程是________________。5.方程$x^2-(m-1)x+2=0$两根之和为$3$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)则$\(m=\)$________。##四\(x_1、x_2\)解答题1.已知$x_1,x_2$是方程$x^2-2x-1=0$的两个实数根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)不解方程求：（1）$x_1+x_2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1x_2$；（2）$(x_1+1)(x_2+1)$2.已知一元二次方程$2x^2-5x+1=0$两根为$x_1,x_2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求$(x_1-x_2)^2$的值。3.已知关于$x$的一元二次方程$x^2-4x+k+1=0$有两个实数根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)且$x_1x_2=2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求$k$的值。4.拓展探究：已知关于$x$的方程$x^2-2(m+1)x+m^2-3=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)若$x_1+x_2=x_1x_2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求$m$的值。##五\(x_1、x_2\)参考答案与解析###选择题1.A。解析：$x_1+x_2=-\dfrac{b}{a}=4$。2.D。解析：$x_1x_2=\dfrac{c}{a}=-\dfrac32$。3.B。解析：两根之积$2\cdotx_2=-6$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)解得$x_2=-3$。4.A。解析：$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=4-2\(m=\)6$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$\(m=\)-1$。5.B。解析：互为倒数\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1x_2=1$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$\(c=1\)$。###填空题1.$-6\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)-2$2.$\dfrac25$；解析：$\dfrac1{x_1}+\dfrac1{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{\frac23}{-\frac53}=-\dfrac25$3.$23$；解析：$x_1+x_2=5\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)x_1x_2=1$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1^2+x_2^2=25-2=23$4.$x^2+x-6=0$；解析：$x^2-(2-3)x+2\times(-3)=0$5.$4$；解析：$x_1+x_2=m-1=3$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$\(m=\)4$###解答题1.解：$a=1,b=-2,c=-1$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1+x_2=2\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)x_1x_2=-1$（1）$x_1+x_2=2\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)x_1x_2=-1$（2）$(x_1+1)(x_2+1)=x_1x_2+x_1+x_2+1=-1+2+1=2$2.解：$a=2,b=-5,\(c=1\)$$x_1+x_2=\dfrac52\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)x_1x_2=\dfrac12$$(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=\left(\dfrac52\right)^2-4\times\dfrac12=\dfrac{25}{4}-2=\dfrac{17}{4}$3.解：$x_1x_2=k+1=2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)得$\(k=1\)$。检验：方程$x^2-4x+2=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$\Delta=16-8=8>0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)存在实数根。$\therefore\(k=1\)$。4.解：$x_1+x_2=2(m+1)$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1x_2=m^2-3$由题意：$2(m+1)=m^2-3$整理：$m^2-2m-5=0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)解得$m_1=1+\sqrt6\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)m_2=1-\sqrt6$。检验判别式$\Delta=4(m+1)^2-4(m^2-3)=8m+16\ge0$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$m\ge-2$。$1+\sqrt6>-2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$1-\sqrt6>-2$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)两个解都满足条件。$\therefore\(m=\)1+\sqrt6$或$\(m=\)1-\sqrt6$。##六\(x_1、x_2\)易错总结韦达定理最核心的限制条件：**只有方程有实数根（$\Delta\ge0$）\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)根与系数关系才能使用**。含参数的大题\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)求出参数后\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)必须带回检验判别式\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)很多同学直接忽略这一步\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)直接丢分。公式记忆：两根之和是$-\dfrac{b}{a}$\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)负号极易遗忘\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)是高频错误。代数式变形一定要熟练\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)$x_1^2+x_2^2$\(x_1、x_2\)$(x_1-x_2)^2$\(x_1、x_2\)$\dfrac1{x_1}+\dfrac1{x_2}$这三类变形是考试必考。看到两根互为相反数：$x_1+x_2=0$；两根互为倒数：$x_1x_2=1$；两根一正一负：$x_1x_2<0$。已知一个根求另一个根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)既可以代入求参数再解方程\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)也可以直接用韦达定理\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)后者计算更快。构造方程记住格式：$x^2-(两根之和)x+两根之积=0$。注意\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)韦达定理不能用来求根\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)只是反映根与系数之间的数量关系。综合大题经常把判别式和韦达定理放在一起\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)先利用$\Delta$锁定参数范围\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)再用韦达定理计算代数式的值\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)一定要注意先后顺序。做题时\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)先看题目是否为一元二次方程\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\sqrt6，m_2=1-\sqrt6\)x_1x_2=\dfrac12\)x_1x_2=-1\)x_1x_2=1\)-2\)x_1x_2=q\)二次项系数含参数时\(x_1+x_2=-p\(-6\(x_1+x_2=5\(x_1+x_2=2\(x_1+x_2=\dfrac52\(m_1=1+\