初中九年级数学《直接开平方法解一元二次方程》知识清单

初中九年级数学《直接开平方法解一元二次方程》知识清单一、学习目标与核心素养定位（一）学习目标1.【基础】理解解一元二次方程的基本思想是“降次”，即把二次方程转化为一次方程。2.【核心】掌握直接开平方法的理论依据——平方根的定义，能够准确判断一个一元二次方程是否适合用直接开平方法求解。3.【技能】熟练运用直接开平方法解形如x2=px^2=px2=p或(mx+n)2=p(mx+n)^2=p(mx+n)2=p（其中p≥0p\ge0p≥0，m≠0m\neq0m=0）的一元二次方程，准确求出方程的根。4.【难点】理解并掌握当p<0p<0p<0时，方程在实数范围内无解的情况，并能用规范的数学语言进行表述。（二）核心素养培养1.【数学抽象】从求平方根的运算中抽象出解一元二次方程的方法，体会数学知识之间的内在联系。2.【逻辑推理】通过分析方程结构特征，推导出使用直接开平方法的条件，培养有条理的思考能力和推理能力。3.【数学运算】在开平方运算过程中，准确处理符号、系数及根式的化简，提升运算的准确性和严谨性。4.【★数学建模】能将实际问题（如面积问题、动点问题中的等量关系）抽象为符合(mx+n)2=p(mx+n)^2=p(mx+n)2=p形式的数学模型，并用直接开平方法解决。二、核心概念与原理溯源（一）一元二次方程解法核心思想——降次1.【重要】我们已经学习了一元一次方程的解法，其本质是通过“合并同类项”、“移项”、“系数化为1”等步骤，将方程转化为x=ax=ax=a的形式。而一元二次方程含有x2x^2x2项，未知数的最高次数是2。因此，解一元二次方程的核心思想就是“降次”，即将二次方程转化为一次方程来求解。2.【方法】“降次”的途径是多样的，直接开平方法、配方法、公式法、因式分解法等都是实现“降次”的具体策略。本章我们将逐一学习这些方法，而直接开平方法是最基础、最本源的一种。（二）直接开平方法的理论基石——平方根的定义1.【★基础】平方根的定义：如果一个数的平方等于aaa，那么这个数叫做aaa的平方根。用数学式子表达为：若x2=ax^2=ax2=a，则xxx是aaa的平方根。2.【要点】根据平方根的定义，我们可以得出以下重要结论：1.3.当a>0a>0a>0时，aaa有两个平方根，它们互为相反数，记作x=±ax=\pm\sqrt{a}x=±a<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">​。2.4.当a=0a=0a=0时，aaa有一个平方根，即000，记作x=0x=0x=0。3.5.【★难点】当a<0a<0a<0时，在实数范围内，负数没有平方根。因此，不存在实数xxx使得x2=ax^2=ax2=a成立。6.【原理】直接开平方法正是逆向运用了平方根的定义。当我们把一个一元二次方程最终化成形如“一个含未知数的式子的平方=一个常数”的形式时，就可以直接通过求这个常数的平方根，来达到“降次”并求解未知数的目的。三、直接开平方法的标准模型与解题步骤（一）【▲高频考点】标准模型一：x2=px^2=px2=p型1.【特征】方程左边是未知数的完全平方，且系数为1；右边是一个常数ppp。2.【解题步骤】1.3.第一步（判号）：判断常数ppp的符号。2.4.第二步（求解）：1.3.5.若p>0p>0p>0，则根据平方根的定义，直接开平方得x=±px=\pm\sqrt{p}x=±p<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">​。即方程有两个不相等的实数根：x1=px_1=\sqrt{p}x1​=p<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=−px_2=\sqrt{p}x2​=−p<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">​。2.4.6.若p=0p=0p=0，则x2=0x^2=0x2=0，开平方得x=0x=0x=0。即方程有两个相等的实数根（我们常记作x1=x2=0x_1=x_2=0x1​=x2​=0）。3.5.7.【★易错点】若p<0p<0p<0，则方程在实数范围内无解（或称为无实数根）。记作“原方程无实数根”。注意，不能说“方程无解”，因为在后续学习复数概念后，它是有解的，现阶段我们特指“无实数根”。8.【示例】解方程4x2−9=04x^29=04x2−9=0。1.9.【分析】先将方程化为x2=px^2=px2=p的形式。移项得4x2=94x^2=94x2=9，两边同除以4得x2=94x^2=\frac{9}{4}x2=49​。2.10.【解答】因为94>0\frac{9}{4}>049​>0，所以x=±94=±32x=\pm\sqrt{\frac{9}{4}}=\pm\frac{3}{2}x=±49​<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">​=±23​。3.11.【结论】原方程的解为x1=32x_1=\frac{3}{2}x1​=23​，x2=−32x_2=\frac{3}{2}x2​=−23​。（二）【▲▲高频考点】标准模型二：(mx+n)2=p(mx+n)^2=p(mx+n)2=p型1.【特征】方程左边是含有未知数的一次二项式的完全平方（m≠0m\neq0m=0），右边是一个常数ppp。这是直接开平方法最核心、最常见的考查形式。2.【解题步骤】1.3.第一步（标准化）：确保方程已经转化为(mx+n)2=p(mx+n)^2=p(mx+n)2=p的标准形式。如果右边不为0，或左边不是完全平方，需先通过移项、合并、系数变形等手段进行整理。2.4.第二步（判号）：判断常数ppp的符号。3.5.第三步（开方降次）：1.4.6.若p≥0p\ge0p≥0，则对等式两边同时开平方，得到mx+n=±pmx+n=\pm\sqrt{p}mx+n=±p<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">​。这一步将一元二次方程降次为两个一元一次方程。2.5.7.【★重要】这里的“±\pm±”不能遗漏，它保证了方程根的完备性。3.6.8.若p<0p<0p<0，则直接判定原方程无实数根。7.9.第四步（解一次方程）：分别解这两个一元一次方程mx+n=pmx+n=\sqrt{p}mx+n=p<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">​和mx+n=−pmx+n=\sqrt{p}mx+n=−p<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">​，得到原方程的两个根x1x_1x1​和x2x_2x2​。8.10.第五步（根的化简）：如果求得的根含有根号，需要化为最简二次根式；如果能合并同类项或分母有理化，也应一并完成。11.【示例】解方程3(x−1)2−6=03(x1)^26=03(x−1)2−6=0。1.12.【分析】首先将方程化为标准形式。移项得3(x−1)2=63(x1)^2=63(x−1)2=6，两边同除以3得(x−1)2=2(x1)^2=2(x−1)2=2。2.13.【解答】因为2>02>02>0，直接开平方得x−1=±2x1=\pm\sqrt{2}x−1=±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">​。3.14.解一次方程：由x−1=2x1=\sqrt{2}x−1=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">​得x1=1+2x_1=1+\sqrt{2}x1​=1+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">​；由x−1=−2x1=\sqrt{2}x−1=−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">​得x2=1−2x_2=1\sqrt{2}x2​=1−2<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">​。4.15.【结论】原方程的解为x1=1+2x_1=1+\sqrt{2}x1​=1+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=1−2x_2=1\sqrt{2}x2​=1−2<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">​。16.【示例】解方程(2x+3)2=−5(2x+3)^2=5(2x+3)2=−5。1.17.【解答】因为−5<05<0−5<0，而任何实数的平方都不可能为负数，所以原方程无实数根。四、【▲难点】直接开平方法的进阶应用与变形（一）系数不为1的转化1.【考点】当方程形式为a(mx+n)2=ca(mx+n)^2=ca(mx+n)2=c（a≠0a\neq0a=0）时，不能直接开平方。必须先将系数化为1，即两边同时除以aaa，得到(mx+n)2=ca(mx+n)^2=\frac{c}{a}(mx+n)2=ac​的形式后，再按照标准模型求解。2.【关键】注意ca\frac{c}{a}ac​的符号判断，这是决定方程是否有实数根的关键。3.【示例】解方程2(3x+1)2=82(3x+1)^2=82(3x+1)2=8。1.4.【正解】两边同除以2得(3x+1)2=4(3x+1)^2=4(3x+1)2=4。因为4>04>04>0，开平方得3x+1=±23x+1=\pm23x+1=±2。2.5.当3x+1=23x+1=23x+1=2时，3x=13x=13x=1，x1=13x_1=\frac{1}{3}x1​=31​。3.6.当3x+1=−23x+1=23x+1=−2时，3x=−33x=33x=−3，x2=−1x_2=1x2​=−1。（二）含根式的方程1.【考点】方程中出现二次根式，解法步骤不变，但要注意化简和分母有理化。2.【示例】解方程(x−2)2=3(x\sqrt{2})^2=3(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=3。1.3.【解答】开平方得x−2=±3x\sqrt{2}=\pm\sqrt{3}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">​=±3<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">​。2.4.所以x1=2+3x_1=\sqrt{2}+\sqrt{3}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">​+3<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=2−3x_2=\sqrt{2}\sqrt{3}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">​−3<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">​。（三）含参数的方程1.【★难点】当方程中的常数被字母参数替代时，需要对参数进行分类讨论。2.【示例】解关于xxx的方程(x−1)2=k(x1)^2=k(x−1)2=k。1.3.【分析】参数kkk的符号不确定，必须分类讨论。2.4.【解答】1.3.5.当k>0k>0k>0时，原方程有两个不相等的实数根：x=1±kx=1\pm\sqrt{k}x=1±k<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">​。2.4.6.当k=0k=0k=0时，原方程有两个相等的实数根：x1=x2=1x_1=x_2=1x1​=x2​=1。3.5.7.当k<0k<0k<0时，原方程无实数根。8.【示例】解关于xxx的方程ax2−b=0ax^2b=0ax2−b=0（a≠0a\neq0a=0）。1.9.【分析】移项得ax2=bax^2=bax2=b，x2=bax^2=\frac{b}{a}x2=ab​。此时需要对ba\frac{b}{a}ab​的符号进行讨论。2.10.【解答】1.3.11.若a,ba,ba,b同号或b=0b=0b=0，即ba≥0\frac{b}{a}\ge0ab​≥0，则x=±bax=\pm\sqrt{\frac{b}{a}}x=±ab​<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">​。2.4.12.若a,ba,ba,b异号，即ba<0\frac{b}{a}<0ab​<0，则原方程无实数根。五、直接开平方法与配方法的关联（承上启下）1.【思想铺垫】直接开平方法只能解决左边是完全平方式，右边是常数的方程。对于一般形式ax2+bx+c=0ax^2+bx+c=0ax2+bx+c=0，我们无法直接开平方。2.【方法过渡】配方法的核心思想就是通过“配方”，将任意一个一元二次方程恒等变形为(mx+n)2=p(mx+n)^2=p(mx+n)2=p的形式，然后再利用直接开平方法求解。因此，直接开平方法是配方法乃至整个公式法的基础和最后一步。3.【内在联系】理解直接开平方法，是理解更高阶解法的钥匙。例如，对于方程x2+2x−3=0x^2+2x3=0x2+2x−3=0，配方后得(x+1)2=4(x+1)^2=4(x+1)2=4，接下来就是应用直接开平方法求解。六、【必考】常见题型与解题策略（一）题型一：判断方程根的个数及情况1.【考查方式】给出几个方程，判断哪些可以用直接开平方法求解，或者判断用直接开平方法求解的方程根的情况。2.【解题策略】紧扣标准形式(mx+n)2=p(mx+n)^2=p(mx+n)2=p。关键是看方程能否化为左边是含未知数式子的平方，右边是非负常数的形式。3.【示例】下列方程中，有实数根的是（）A.x2+1=0x^2+1=0x2+1=0B.(x−1)2=−1(x1)^2=1(x−1)2=−1C.(2x+3)2=0(2x+3)^2=0(2x+3)2=0D.(x+2)2+1=0(x+2)^2+1=0(x+2)2+1=01.4.【解析】A项化为x2=−1x^2=1x2=−1，无实数根；B项(x−1)2=−1(x1)^2=1(x−1)2=−1，无实数根；C项(2x+3)2=0(2x+3)^2=0(2x+3)2=0，得2x+3=02x+3=02x+3=0，有实数根x=−32x=\frac{3}{2}x=−23​；D项化为(x+2)2=−1(x+2)^2=1(x+2)2=−1，无实数根。故选C。（二）题型二：解标准形式的一元二次方程1.【考查方式】直接给出形如x2=ax^2=ax2=a或(x+b)2=c(x+b)^2=c(x+b)2=c的方程，要求写出解。2.【解题策略】严格按照“判号—开方—求解”三步走，注意“±\pm±”不能丢，结果要化简。3.【示例】方程(x−2)2=9(x2)^2=9(x−2)2=9的解是______。1.4.【解析】x−2=±3x2=\pm3x−2=±3，x=2±3x=2\pm3x=2±3，所以x1=5x_1=5x1​=5，x2=−1x_2=1x2​=−1。答案：x1=5x_1=5x1​=5，x2=−1x_2=1x2​=−1。（三）题型三：解需要变形的方程（高频）1.【考查方式】方程不是直接给出的标准形式，需要先进行移项、合并同类项、去括号、系数化为1等操作。2.【解题策略】化归思想。将所有步骤的目标都指向化为(mx+n)2=p(mx+n)^2=p(mx+n)2=p的形式。3.【示例】一元二次方程4x2−4x+1=04x^24x+1=04x2−4x+1=0的解为______。1.4.【分析】观察左边，4x2−4x+1=(2x−1)24x^24x+1=(2x1)^24x2−4x+1=(2x−1)2。2.5.【解答】原方程即(2x−1)2=0(2x1)^2=0(2x−1)2=0，开平方得2x−1=02x1=02x−1=0，解得x1=x2=12x_1=x_2=\frac{1}{2}x1​=x2​=21​。6.【示例】解方程9(x+1)2−16(x−1)2=09(x+1)^216(x1)^2=09(x+1)2−16(x−1)2=0。1.7.【分析】这是一个较为复杂的方程，可以通过移项构造平方形式。2.8.【解法一】移项得9(x+1)2=16(x−1)29(x+1)^2=16(x1)^29(x+1)2=16(x−1)2。两边同除以（注意x=1x=1x=1是否为解？可先判断）观察，两边都是非负，可以直接开平方。开方得3∣x+1∣=4∣x−1∣3|x+1|=4|x1|3∣x+1∣=4∣x−1∣，此方法需讨论绝对值，较繁琐。3.9.【解法二】（推荐）利用平方差公式。移项得9(x+1)2−16(x−1)2=09(x+1)^216(x1)^2=09(x+1)2−16(x−1)2=0，即[3(x+1)]2−[4(x−1)]2=0[3(x+1)]^2[4(x1)]^2=0[3(x+1)]2−[4(x−1)]2=0。利用平方差公式分解得[3(x+1)+4(x−1)]⋅[3(x+1)−4(x−1)]=0[3(x+1)+4(x1)]\cdot[3(x+1)4(x1)]=0[3(x+1)+4(x−1)]⋅[3(x+1)−4(x−1)]=0，即(3x+3+4x−4)(3x+3−4x+4)=0(3x+3+4x4)(3x+34x+4)=0(3x+3+4x−4)(3x+3−4x+4)=0，化简得(7x−1)(−x+7)=0(7x1)(x+7)=0(7x−1)(−x+7)=0，解得x1=17x_1=\frac{1}{7}x1​=71​，x2=7x_2=7x2​=7。（四）题型四：与几何或其他学科的综合应用1.【考查方式】在几何图形的面积、边长计算，或物理中的自由落体运动、动点问题中，列出方程后，需要用直接开平方法求解。2.【解题策略】关键是根据题意正确列出方程，然后对方程进行化简，看是否适合用直接开平方法。通常在得到形如（边长）2^22=面积的方程时，此法非常简便。3.【示例】一个圆的面积是16π16\pi16πcm2^22，求这个圆的半径。1.4.【解析】设圆的半径为rrrcm，根据面积公式有πr2=16π\pir^2=16\piπr2=16π。两边同除以π\piπ得r2=16r^2=16r2=16。因为r>0r>0r>0，直接开平方得r=4r=4r=4（负值舍去）。所以圆的半径为4cm。（五）题型五：新定义与阅读理解题1.【考查方式】给出一段材料，介绍一种新的运算规则，然后要求根据规则解方程。这种题型常将开平方运算与新的符号或规则结合。2.【解题策略】认真阅读材料，理解新规则的本质，将其转化为我们熟悉的数学模型，通常是转化为(mx+n)2=p(mx+n)^2=p(mx+n)2=p的形式。七、【★致命】易错点与避坑指南1.【易错点一】忘掉“±\pm±”1.2.【错误表现】解方程x2=4x^2=4x2=4，直接解得x=2x=2x=2，遗漏了x=−2x=2x=−2。2.3.【正确做法】牢记“正数的平方根有两个，它们互为相反数”，开方后必须写上“±\pm±”。对于(mx+n)2=p(p>0)(mx+n)^2=p(p>0)(mx+n)2=p(p>0)的形式，开方后得到mx+n=±pmx+n=\pm\sqrt{p}mx+n=±p<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">​。...4.【避坑技巧】解题时，可将开方步骤在心里默念为“等于正负根号下...”。5.【易错点二】忽略p<0p<0p<0的情况1.6.【错误表现】解方程(x−1)2=−3(x1)^2=3(x−1)2=−3，开方得x−1=±−3x1=\pm\sqrt{3}x−1=±−3<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">​，然后在实数范围内继续求解，得出错误答案。2.7.【正确做法】在进行开平方运算前，必须首先检查右边常数ppp的符号。一旦发现p<0p<0p<0，立即下结论“原方程无实数根”，后续步骤无需进行。3.8.【避坑技巧】养成“先判号，后开方”的良好解题习惯。将判号作为开平方操作的必要前置步骤。9.【易错点三】系数处理不当1.10.【错误表现】解方程2x2=82x^2=82x2=8，直接开平方得2x=±82x=\pm\sqrt{8}2x=±8<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">​，然后得到x=±2x=\pm\sqrt{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">​。这种错误在于没有先将x2x^2x2的系数化为1。2.11.【正确做法】必须先将方程化为x2=px^2=px2=p或(mx+n)2=p(mx+n)^2=p(mx+n)2=p的形式。对于2x2=82x^2=82x2=8，应两边同除以2得x2=4x^2=4x2=4，再开平方得x=±2x=\pm2x=±2。3.12.【避坑技巧】任何非1的系数，都必须通过除法运算化为1之后，才能对“平方”进行开方操作。13.【易错点四】对“p=0p=0p=0”的理解偏差1.14.【错误表现】解方程(x−3)2=0(x3)^2=0(x−3)2=0，开方得x−3=0x3=0x−3=0，然后只写了一个根x=3x=3x=3。2.15.【正确做法】虽然两个根相等，但在答题规范上，通常需要写成x1=x2=3x_1=x_2=3x1​=x2​=3的形式，以体现“两个相等的实数根”这一概念，也符合一元二次方程“有两个根”的代数基本定理的要求。3.16.【避坑技巧】理解“平方等于0”意味着这个式子本身就是0，解出的根虽然数值相同，但在计数上视为两个。17.【易错点五】结果未化简1.18.【错误表现】解方程x2=8x^2=8x2=8，得到x=±8x=\pm\sqrt{8}x=±8<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">​后就作为最终答案。2.19.【正确做法】必须将二次根式化为最简形式，即x=±22x=\pm2\sqrt{2}x=±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">​。3.20.【避坑技巧】复习最简二次根式的两个条件：（1）被开方数不含分母；（2）被开方数中不含能开得尽方的因数或因式。解出带根号的根后，务必检查化简。21.【易错点六】移项、合并同类项时符号错误1.22.【错误表现】解方程(x+2)2+3=0(x+2)^2+3=0(x+2)2+3=0，移项得(x+2)2=3(x+2)^2=3(x+2)2=3。2.23.【正确做法】移项要变号。(x+2)2+3=0(x+2)^2+3=0(x+2)2+3=0移项应为(x+2)2=−3(x+2)^2=3(x+2)2=−3。3.24.【避坑技巧】移项时，默念“过河要拆桥，过等号要变号”。或者直接在等式两边同时加上或减去同一个数（式）。八、思维拓展与一题多解（一）换元思想在直接开平方法中的渗透1.【思想】对于一些结构复杂的方程，我们可以将其中某一部分看成一个整体（设为新未知数），从而简化方程形式，使之符合(mx+n)2=p(mx+n)^2=p(mx+n)2=p的模型。2.【示例】解方程(x2−2x)2−2(x2−2x)−3=0(x^22x)^22(x^22x)3=0(x2−2x)2−2(x2−2x)−3=0。1.3.【分析】这个方程看起来是四次方程，但如果我们把x2−2xx^22xx2−2x看成一个整体，设为yyy，则原方程变为y2−2y−3=0y^22y3=0y2−2y−3=0。这虽然是一个关于yyy的一元二次方程，但形式上不是平方式。我们可以先解这个关于yyy的方程。分解因式得(y−3)(y+1)=0(y3)(y+1)=0(y−3)(y+1)=0，所以y=3y=3y=3或y=−1y=1y=−1。2.4.【回归】当y=3y=3y=3时，x2−2x=3x^22x=3x2−2x=3，即x2−2x−3=0x^22x3=0x2−2x−3=0，可以配方得(x−1)2=4(x1)^2=4(x−1)2=4，开平方得x−1=±2x1=\pm2x−1=±2，解得x1=3,x2=−1x_1=3,x_2=1x1​=3,x2​=−1。3.5.当y=−1y=1y=−1时，x2−2x=−1x^22x=1x2−2x=−1，即x2−2x+1=0x^22x+1=0x2−2x+1=0，配方得(x−1)2=0(x1)^2=0(x−1)2=0，解得x3=x4=1x_3=x_4=1x3​=x4​=1。4.6.【结论】原方程的解为x1=3,x2=−1,x3=x4=1x_1=3,x_2=1,x_3=x_4=1x1​=3,x2​=−1,x3​=x4​=1。5.7.【拓展】这里虽然最终没有直接使用一次换元后的直接开平方法，但换元思想简化了问题，而中间步骤依然用到了直接开平方法来解(x−1)2=4(x1)^2=4(x−1)2=4和(x−1)2=0(x1)^2=0(x−1)2=0。（二）数形结合理解直接开平方法1.【几何意义】方程x2=p(p>0)x^2=p(p>0)x2=p(p>0)的解，在几何上可以理解为：在平面直角坐标系中，求抛物线y=x2y=x^2y=x2与直线y=py=py=p的交点横坐标。这两个交点关于y轴对称，其横坐标恰好互为相反数，即x=±px=\pm\sqrt{p}x=±p<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.【几何意义】方程(x−h)2=p(p>0)(xh)^2=p(p>0)(x−h)2=p(p>0)的解，对应着抛物线y=(x−h)2y=(xh)^2y=(x−h)2与直线y=py=py=p的交点横坐标。抛物线的顶点在(h,0)(h,0)(h,0)，开口向上，与水平线y=py=py=p相交于两点，这两点关于直线x=hx=hx=h对称，其横坐标为h±ph\pm\sqrt{p}h±p<pathd="M95,702c2.7,0,7.17,2.7,13.5,8c5.8,5.3,9.5,10,9.5,14c0,2,0.3,3.3,1,4c1.3,2.7,23.83,20.7,67.5,54c44.2,33.3,65.8,50.3,66.5,51c1.3,1.3,3,2,5,2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,71,104,213c68.7,142,137.5,285,206.5,429c69,1