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九年级数学解答题一般
题目
用适当的方法求解下列方程:
(1)x22x3=0(1)x^{2}-2x-3=0
(2)2x2+3x1=0(2)2x^{2}+3x-1=0
(3)x2=6x1(3)x^{2}=6x-1
(4)(x2)2=3(x2)(4)\left(x-2\right)^{2}=3\left(x-2\right).
知识点:运用公式法、解一元二次方程——直接开平方法、解一元二次方程——因式分解法章节:第21章 一元二次方程 / 21.2 一元二次方程的解法 / 21.2.1 特殊的一元二次方程的解法

答案与解析

答案

(1)(x3)(x+1)=0\left(1\right)\left(x-3\right)\left(x+1\right)=0
x3=0x-3=0x+1=0x+1=0
所以x1=3x_{1}=3x2=1x_{2}=-1
(2)2x2+3x1=0(2)2x^{2}+3x-1=0
a=2\because a=2b=3b=3c=1c=-1
Δ=324×2×(1)=17\therefore \Delta =3^{2}-4\times 2\times \left(-1\right)=17
x=3±172×2\therefore x=\frac{-3±\sqrt{17}}{2×2}
x1=3+174\therefore x_{1}=\frac{-3+\sqrt{17}}{4}x2=3174x_{2}=\frac{-3-\sqrt{17}}{4}.
(3)x2=6x1(3)x^{2}=6x-1
x26x=1x^{2}-6x=-1
x26x+9=1+9x^{2}-6x+9=-1+9
(x3)2=8(x-3)^{2}=8
x3=±22x-3=\pm 2\sqrt{2}
所以x1=3+22x_{1}=3+2\sqrt{2}x2=322x_{2}=3-2\sqrt{2}
(4)(x2)2=3(x2)(4)\left(x-2\right)^{2}=3\left(x-2\right)
(x2)23(x2)=0(x-2)^{2}-3\left(x-2\right)=0
(x2)(x23)=0(x-2)\left(x-2-3\right)=0
x2=0x-2=0x23=0x-2-3=0
所以x1=2x_{1}=2x2=5x_{2}=5.

解析

(1)(x3)(x+1)=0\left(1\right)\left(x-3\right)\left(x+1\right)=0
x3=0x-3=0x+1=0x+1=0
所以x1=3x_{1}=3x2=1x_{2}=-1
(2)2x2+3x1=0(2)2x^{2}+3x-1=0
a=2\because a=2b=3b=3c=1c=-1
Δ=324×2×(1)=17\therefore \Delta =3^{2}-4\times 2\times \left(-1\right)=17
x=3±172×2\therefore x=\frac{-3±\sqrt{17}}{2×2}
x1=3+174\therefore x_{1}=\frac{-3+\sqrt{17}}{4}x2=3174x_{2}=\frac{-3-\sqrt{17}}{4}.
(3)x2=6x1(3)x^{2}=6x-1
x26x=1x^{2}-6x=-1
x26x+9=1+9x^{2}-6x+9=-1+9
(x3)2=8(x-3)^{2}=8
x3=±22x-3=\pm 2\sqrt{2}
所以x1=3+22x_{1}=3+2\sqrt{2}x2=322x_{2}=3-2\sqrt{2}
(4)(x2)2=3(x2)(4)\left(x-2\right)^{2}=3\left(x-2\right)
(x2)23(x2)=0(x-2)^{2}-3\left(x-2\right)=0
(x2)(x23)=0(x-2)\left(x-2-3\right)=0
x2=0x-2=0x23=0x-2-3=0
所以x1=2x_{1}=2x2=5x_{2}=5.

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