初中九年级数学 配方法解一元二次方程（第2课时）知识清单

初中九年级数学配方法解一元二次方程（第2课时）知识清单【知识框架构建】本课时在掌握配方法基本步骤的基础上，重点解决二次项系数不为1的一元二次方程的配方求解，并深入探究配方法背后的代数恒等变形思想。核心内容涵盖从一般式到顶点式的转化、无理数系数的处理、以及配方法在实际问题建模中的应用。一、【基础巩固与概念深化】（一）配方法的核心原理【重要】配方法解一元二次方程的本质是：通过恒等变形，将一般形式的一元二次方程转化为形如(x+m)2=n(x+m)^2=n(x+m)2=n的方程，然后利用直接开平方法求解。这一过程体现了数学中的“化归”思想，即将未知的、复杂的问题转化为已知的、简单的问题。（二）本课时核心知识图谱1.二次项系数不为1的配方步骤【高频考点】1.步骤一：系数化“1”。将方程两边同时除以二次项系数，使二次项系数变为1。2.步骤二：移项。将常数项移到方程的右边。3.步骤三：配方。方程两边同时加上一次项系数一半的平方。4.步骤四：变形。将方程左边写成完全平方式，右边合并常数。5.步骤五：开方。根据平方根的意义，若右边为非负数，则方程有两个实数根；若为负数，则方程无实数根。6.步骤六：求解。解两个一元一次方程，得出原方程的根。1.配方的基本模型【非常重要】对于二次三项式ax2+bx+cax^2+bx+cax2+bx+c（a≠0a\neq0a=0），配方的通用模型为：ax2+bx+c=a(x2+bax)+c=a[x2+bax+(b2a)2]+c−a×(b2a)2=a(x+b2a)2+4ac−b24aax^2+bx+c=a\left(x^2+\frac{b}{a}x\right)+c=a\left[x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2\right]+ca\times\left(\frac{b}{2a}\right)^2=a\left(x+\frac{b}{2a}\right)^2+\frac{4acb^2}{4a}ax2+bx+c=a(x2+ab​x)+c=a[x2+ab​x+(2ab​)2]+c−a×(2ab​)2=a(x+2ab​)2+4a4ac−b2​该模型揭示了二次函数y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c的顶点坐标为(−b2a,4ac−b24a)\left(\frac{b}{2a},\frac{4acb^2}{4a}\right)(−2ab​,4a4ac−b2​)，这也是配方法在后续学习二次函数中的重要应用。（三）必备知识与技能诊断1.前置知识回顾1.完全平方公式：(a±b)2=a2±2ab+b2(a\pmb)^2=a^2\pm2ab+b^2(a±b)2=a2±2ab+b2。必须能熟练正向与逆向运用。2.分式的基本性质与运算：处理系数为分数或无理数时的通分与化简。3.平方根的性质：一个正数有两个平方根，它们互为相反数；0的平方根是0；负数没有平方根。1.本课时核心能力点1.代数变形能力：能够正确处理系数化“1”后出现的分数。2.符号感：在配方过程中，准确判断和添加“一次项系数一半的平方”，尤其是当一次项系数为负数时。3.判别意识：在开平方前，养成先判断右边常数项符号的习惯，从而预判方程根的情况。二、【核心方法与解题策略】（一）二次项系数为1与不为1的对比分析【难点】处理类型二次项系数为1二次项系数不为1示例方程x2−6x+8=0x^26x+8=0x2−6x+8=02x2−5x−3=02x^25x3=02x2−5x−3=0第一步直接移项：x2−6x=−8x^26x=8x2−6x=−8化系数为1：x2−52x−32=0x^2\frac{5}{2}x\frac{3}{2}=0x2−25​x−23​=0第二步配方：两边加(−3)2=9(3)^2=9(−3)2=9移项：x2−52x=32x^2\frac{5}{2}x=\frac{3}{2}x2−25​x=23​第三步(x−3)2=1(x3)^2=1(x−3)2=1配方：两边加(54)2=2516\left(\frac{5}{4}\right)^2=\frac{25}{16}(45​)2=1625​第四步x−3=±1x3=\pm1x−3=±1(x−54)2=32+2516=2416+2516=4916\left(x\frac{5}{4}\right)^2=\frac{3}{2}+\frac{25}{16}=\frac{24}{16}+\frac{25}{16}=\frac{49}{16}(x−45​)2=23​+1625​=1624​+1625​=1649​第五步x1=4,x2=2x_1=4,x_2=2x1​=4,x2​=2x−54=±74x\frac{5}{4}=\pm\frac{7}{4}x−45​=±47​，得x1=3,x2=−12x_1=3,x_2=\frac{1}{2}x1​=3,x2​=−21​【易错警示】在系数化“1”时，方程的各项都要除以二次项系数，不能遗漏常数项。配方时，所加的数是一次项系数（化简后的）一半的平方，而非原系数的平方。（二）特殊类型方程的配方技巧1.含有无理数系数的方程例：解方程2x2−3x+2=0\sqrt{2}x^23x+\sqrt{2}=02<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​x2−3x+2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=0。策略：同样遵循“化系数为1”的原则，两边同时除以2\sqrt{2}2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​。解：x2−32x+1=0x^2\frac{3}{\sqrt{2}}x+1=0x2−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​3​x+1=0，即x2−322x+1=0x^2\frac{3\sqrt{2}}{2}x+1=0x2−232<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​x+1=0。移项：x2−322x=−1x^2\frac{3\sqrt{2}}{2}x=1x2−232<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​x=−1。配方：一次项系数为−322\frac{3\sqrt{2}}{2}−232<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，其一半为−324\frac{3\sqrt{2}}{4}−432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，平方为(324)2=9×216=98\left(\frac{3\sqrt{2}}{4}\right)^2=\frac{9\times2}{16}=\frac{9}{8}(432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​)2=169×2​=89​。两边加98\frac{9}{8}89​：x2−322x+98=−1+98=18x^2\frac{3\sqrt{2}}{2}x+\frac{9}{8}=1+\frac{9}{8}=\frac{1}{8}x2−232<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​x+89​=−1+89​=81​。即(x−324)2=18\left(x\frac{3\sqrt{2}}{4}\right)^2=\frac{1}{8}(x−432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​)2=81​。开方得：x−324=±122=±24x\frac{3\sqrt{2}}{4}=\pm\frac{1}{2\sqrt{2}}=\pm\frac{\sqrt{2}}{4}x−432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=±22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​1​=±42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。解得：x1=324+24=2x_1=\frac{3\sqrt{2}}{4}+\frac{\sqrt{2}}{4}=\sqrt{2}x1​=432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​+42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​，x2=324−24=22x_2=\frac{3\sqrt{2}}{4}\frac{\sqrt{2}}{4}=\frac{\sqrt{2}}{2}x2​=432<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​−42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。2.含有字母系数的方程（参数问题）【拓展】例：解关于xxx的方程mx2−(2m+1)x+m+2=0mx^2(2m+1)x+m+2=0mx2−(2m+1)x+m+2=0（m≠0m\neq0m=0）。分析：此题为含参数方程，需对参数进行分类讨论。m≠0m\neq0m=0是前提，确保方程为二次方程。解：两边除以mmm：x2−2m+1mx+m+2m=0x^2\frac{2m+1}{m}x+\frac{m+2}{m}=0x2−m2m+1​x+mm+2​=0。移项：x2−2m+1mx=−m+2mx^2\frac{2m+1}{m}x=\frac{m+2}{m}x2−m2m+1​x=−mm+2​。配方：一次项系数为−2m+1m\frac{2m+1}{m}−m2m+1​，一半为−2m+12m\frac{2m+1}{2m}−2m2m+1​，平方为(2m+12m)2\left(\frac{2m+1}{2m}\right)^2(2m2m+1​)2。两边加该平方：x2−2m+1mx+(2m+12m)2=(2m+12m)2−m+2mx^2\frac{2m+1}{m}x+\left(\frac{2m+1}{2m}\right)^2=\left(\frac{2m+1}{2m}\right)^2\frac{m+2}{m}x2−m2m+1​x+(2m2m+1​)2=(2m2m+1​)2−mm+2​。左边为(x−2m+12m)2\left(x\frac{2m+1}{2m}\right)^2(x−2m2m+1​)2，右边通分合并：(2m+1)24m2−4m(m+2)4m2=4m2+4m+1−(4m2+8m)4m2=−4m+14m2=1−4m4m2\frac{(2m+1)^2}{4m^2}\frac{4m(m+2)}{4m^2}=\frac{4m^2+4m+1(4m^2+8m)}{4m^2}=\frac{4m+1}{4m^2}=\frac{14m}{4m^2}4m2(2m+1)2​−4m24m(m+2)​=4m24m2+4m+1−(4m2+8m)​=4m2−4m+1​=4m21−4m​接下来，需要根据1−4m14m1−4m的符号讨论根的情况：1.当1−4m>014m>01−4m>0，即m<14m<\frac{1}{4}m<41​时，方程有两个实数根：x=2m+12m±1−4m2∣m∣x=\frac{2m+1}{2m}\pm\frac{\sqrt{14m}}{2|m|}x=2m2m+1​±2∣m∣1−4m<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。2.当1−4m=014m=01−4m=0，即m=14m=\frac{1}{4}m=41​时，方程有两个相等的实数根：x=2×14+12×14=3212=3x=\frac{2\times\frac{1}{4}+1}{2\times\frac{1}{4}}=\frac{\frac{3}{2}}{\frac{1}{2}}=3x=2×41​2×41​+1​=21​23​​=3。3.当1−4m<014m<01−4m<0，即m>14m>\frac{1}{4}m>41​时，方程无实数根。（三）配方法的逆向应用与代数证明1.证明一个代数式恒为正（或负）【热点】例：求证：无论xxx取何实数，代数式−2x2+4x−52x^2+4x5−2x2+4x−5的值总是负数。证明：将原式配方。−2x2+4x−5=−2(x2−2x)−5=−2[(x−1)2−1]−5=−2(x−1)2+2−5=−2(x−1)2−32x^2+4x5=2(x^22x)5=2[(x1)^21]5=2(x1)^2+25=2(x1)^23−2x2+4x−5=−2(x2−2x)−5=−2[(x−1)2−1]−5=−2(x−1)2+2−5=−2(x−1)2−3∵(x−1)2≥0\because(x1)^2\geq0∵(x−1)2≥0，∴−2(x−1)2≤0\therefore2(x1)^2\leq0∴−2(x−1)2≤0，则−2(x−1)2−3≤−3<02(x1)^23\leq3<0−2(x−1)2−3≤−3<0。故原式的值总是负数。1.求代数式的最值【高频考点】例：求多项式3x2−5x+43x^25x+43x2−5x+4的最小值。解：配方。3x2−5x+4=3(x2−53x)+4=3[x2−53x+(56)2−(56)2]+4=3[(x−56)2−2536]+4=3(x−56)2−2512+4=3(x−56)2+23123x^25x+4=3\left(x^2\frac{5}{3}x\right)+4=3\left[x^2\frac{5}{3}x+\left(\frac{5}{6}\right)^2\left(\frac{5}{6}\right)^2\right]+4=3\left[\left(x\frac{5}{6}\right)^2\frac{25}{36}\right]+4=3\left(x\frac{5}{6}\right)^2\frac{25}{12}+4=3\left(x\frac{5}{6}\right)^2+\frac{23}{12}3x2−5x+4=3(x2−35​x)+4=3[x2−35​x+(65​)2−(65​)2]+4=3[(x−65​)2−3625​]+4=3(x−65​)2−1225​+4=3(x−65​)2+1223​∵3(x−56)2≥0\because3\left(x\frac{5}{6}\right)^2\geq0∵3(x−65​)2≥0，∴\therefore∴当x=56x=\frac{5}{6}x=65​时，原式有最小值，最小值为2312\frac{23}{12}1223​。【总结】通过配方将二次三项式转化为a(x−h)2+ka(xh)^2+ka(x−h)2+k的形式，其最值由aaa的符号和kkk决定：若a>0a>0a>0，则最小值为kkk；若a<0a<0a<0，则最大值为kkk。三、【考点、考向与题型全析】（一）高频考点与考查方式1.直接考查配方法解方程（基础题）1.题型：解方程3x2−6x+2=03x^26x+2=03x2−6x+2=0。2.考向：考查基本步骤的准确性，特别是系数化“1”后的配方计算。1.配方过程的选择与填空（易错题）1.题型：用配方法解方程2x2−4x−1=02x^24x1=02x2−4x−1=0时，方程两边都应加上______。2.解析：先化系数为1得x2−2x=12x^22x=\frac{1}{2}x2−2x=21​，应加上一次项系数一半的平方，即(−1)2=1(1)^2=1(−1)2=1。故应加“1”。3.陷阱：部分学生容易忽略第一步，直接在原方程两边加(−4/2)2=4(4/2)^2=4(−4/2)2=4，导致错误。1.与完全平方式结合的综合题【重要】1.题型：若关于xxx的二次三项式x2+2(m−3)x+25x^2+2(m3)x+25x2+2(m−3)x+25是完全平方式，求mmm的值。2.解析：完全平方式意味着它可以写成(x±5)2=x2±10x+25(x\pm5)^2=x^2\pm10x+25(x±5)2=x2±10x+25的形式。因此，一次项系数2(m−3)2(m3)2(m−3)必须等于±10\pm10±10。即2(m−3)=102(m3)=102(m−3)=10或2(m−3)=−102(m3)=102(m−3)=−10。解得m=8m=8m=8或m=−2m=2m=−2。3.变式：若二次三项式4x2−12x+m4x^212x+m4x2−12x+m是完全平方式，求mmm的值。4.解析：原式可化为4(x2−3x)+m4(x^23x)+m4(x2−3x)+m。要成为完全平方式，括号内应为(x−32)2=x2−3x+94(x\frac{3}{2})^2=x^23x+\frac{9}{4}(x−23​)2=x2−3x+49​。故原式=4(x−32)2=4x2−12x+94(x\frac{3}{2})^2=4x^212x+94(x−23​)2=4x2−12x+9。所以m=9m=9m=9。1.配方法在代数式恒等变形中的应用（中难题）1.题型：已知a2+b2−4a+8b+20=0a^2+b^24a+8b+20=0a2+b2−4a+8b+20=0，求a+ba+ba+b的值。2.解析：将等式左边分组配方。(a2−4a)+(b2+8b)+20=0⇒(a2−4a+4)+(b2+8b+16)=0⇒(a−2)2+(b+4)2=0(a^24a)+(b^2+8b)+20=0\Rightarrow(a^24a+4)+(b^2+8b+16)=0\Rightarrow(a2)^2+(b+4)^2=0(a2−4a)+(b2+8b)+20=0⇒(a2−4a+4)+(b2+8b+16)=0⇒(a−2)2+(b+4)2=0根据非负数的性质，a−2=0a2=0a−2=0且b+4=0b+4=0b+4=0，得a=2,b=−4a=2,b=4a=2,b=−4。所以a+b=−2a+b=2a+b=−2。1.考向：本题考查配方法与非负数性质的结合，是初中阶段常见的代数求值题型。（二）典型例题精析与规范解答步骤【非常重要】【例1】（基础）用配方法解方程：3x2−6x−5=03x^26x5=03x2−6x−5=0。【规范解答】（1）系数化1：方程两边都除以3，得x2−2x−53=0x^22x\frac{5}{3}=0x2−2x−35​=0。（2）移项：x2−2x=53x^22x=\frac{5}{3}x2−2x=35​。（3）配方：两边同时加(−1)2=1(1)^2=1(−1)2=1，得x2−2x+1=53+1x^22x+1=\frac{5}{3}+1x2−2x+1=35​+1。（4）变形：(x−1)2=83(x1)^2=\frac{8}{3}(x−1)2=38​。（5）开方：x−1=±83=±263x1=\pm\sqrt{\frac{8}{3}}=\pm\frac{2\sqrt{6}}{3}x−1=±38​<pathd="M98390l00c4,6.7,10,10,18,10Hv40H1013.1s83.4,268,264.1,840c180.7,572,277,876.3,289,913c4.7,4.7,12.7,7,24,7s12,0,12,0c1.3,3.3,3.7,11.7,7,25c35.3,125.3,106.7,373.3,214,744c10,12,21,25,33,39s32,39,32,39c6,5.3,15,14,27,26s25,30,25,30c26.7,32.7,52,63,76,91s52,60,52,60s208,722,208,722c56,175.3,126.3,397.3,211,666c84.7,268.7,153.8,488.2,207.5,658.5c53.7,170.3,84.5,266.8,92.5,289.5zMhv40hz">​=±326<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。（6）求解：x1=1+263=3+263x_1=1+\frac{2\sqrt{6}}{3}=\frac{3+2\sqrt{6}}{3}x1​=1+326<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=33+26<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，x2=1−263=3−263x_2=1\frac{2\sqrt{6}}{3}=\frac{32\sqrt{6}}{3}x2​=1−326<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=33−26<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。【例2】（能力）用配方法解方程：2x2−42x+3=02x^24\sqrt{2}x+3=02x2−42<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​x+3=0。【规范解答】（1）系数化1：除以2得，x2−22x+32=0x^22\sqrt{2}x+\frac{3}{2}=0x2−22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​x+23​=0。（2）移项：x2−22x=−32x^22\sqrt{2}x=\frac{3}{2}x2−22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​x=−23​。（3）配方：一次项系数为−222\sqrt{2}−22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​，其一半为−2\sqrt{2}−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​，平方为(2)2=2(\sqrt{2})^2=2(2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​)2=2。两边加2。x2−22x+2=−32+2x^22\sqrt{2}x+2=\frac{3}{2}+2x2−22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​x+2=−23​+2。（4）变形：(x−2)2=12(x\sqrt{2})^2=\frac{1}{2}(x−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​)2=21​。（5）开方：x−2=±22x\sqrt{2}=\pm\frac{\sqrt{2}}{2}x−2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​=±22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。（6）求解：x1=2+22=322x_1=\sqrt{2}+\frac{\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}x1​=2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​+22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=232<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​，x2=2−22=22x_2=\sqrt{2}\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}x2​=2<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​−22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​=22<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,144,104.5,217.7,106.5,221l00c5.3,9.3,12,14,20,14Hv40H845.2724s225.272,467,225.272,467s235,486,235,486c2.7,4.7,9,7,19,7c6,0,10,1,12,3s194,422,194,422s65,47,65,47zM83480Hv40hz">​​。【例3】（应用）已知等腰三角形的两边长a,ba,ba,b是关于xxx的方程x2−(2k+1)x+k2=0x^2(2k+1)x+k^2=0x2−(2k+1)x+k2=0的两个实数根，第三边长为5。当kkk为何值时，△ABC\triangleABC△ABC是等腰三角形？并求出三角形的周长。【思路导航】本题将配方法、一元二次方程根的定义与几何图形结合。需分两种情况讨论：一是a=ba=ba=b（方程有两相等实根）；二是a≠ba\neqba=b，但其中一边与第三边5相等。【规范解答】（1）当a=ba=ba=b时，方程有两个相等的实数根。根的判别式Δ=[−(2k+1)]2−4×1×k2=4k2+4k+1−4k2=4k+1=0\Delta=[(2k+1)]^24\times1\timesk^2=4k^2+4k+14k^2=4k+1=0Δ=[−(2k+1)]2−4×1×k2=4k2+4k+1−4k2=4k+1=0。解得k=−14k=\frac{1}{4}k=−41​。此时方程为x2−(−12+1)x+116=0x^2(\frac{1}{2}+1)x+\frac{1}{16}=0x2−(−21​+1)x+161​=0，即x2−12x+116=0x^2\frac{1}{2}x+\frac{1}{16}=0x2−21​x+161​=0。用配方法解：(x−14)2=0(x\frac{1}{4})^2=0(x−41​)2