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電子課本網(wǎng) 第9頁(yè)

第9頁(yè)

信息發(fā)布者:
我們已經(jīng)學(xué)過(guò)的解一元二次方程的方法有直接開(kāi)平方法、配方法、公式法和因式分解法。
$ 轉(zhuǎn)化過(guò)程:$
1. 方程兩邊同除以$a(a\neq0)$:$x^{2}+\frac{a}x+\frac{c}{a}=0;$
2. 移項(xiàng):$x^{2}+\frac{a}x=-\frac{c}{a};$
3. 配方:兩邊同加$(\frac{2a})^{2},$得$x^{2}+\frac{a}x+(\frac{2a})^{2}=-\frac{c}{a}+(\frac{2a})^{2};$
4. 化為完全平方式:$(x+\frac{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}。$
$注意事項(xiàng):$
1. 化二次項(xiàng)系數(shù)為1時(shí),需確保兩邊同除以$a(a\neq0);$
$2. 移項(xiàng)時(shí)常數(shù)項(xiàng)需變號(hào);$
$3. 配方時(shí)等式兩邊必須同時(shí)加上一次項(xiàng)系數(shù)一半的平方;$
4. 右邊合并同類(lèi)項(xiàng)時(shí)注意通分及符號(hào)運(yùn)算,結(jié)果為$\frac{b^{2}-4ac}{4a^{2}}。$
解: $a = 1,b = 3,c = 2,$
$b2-4ac== 3^{2} - 4 × 1 × 2 = 1,$
$x = \frac{-3 \pm \sqrt{1}}{2 × 1},$
$x_{1} = \frac{-3 + 1}{2} = -1,$$x_{2} = \frac{-3 - 1}{2} = -2;$
解:$a = 1,b = 2,c = -2,$
$b2-4ac== 2^{2} - 4 × 1 × (-2) = 12,$
$x = \frac{-2 \pm \sqrt{12}}{2 × 1} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3},$
$x_{1} = -1 + \sqrt{3},$$x_{2} = -1 - \sqrt{3};$
解:方程兩邊同時(shí)乘以$-1$得:$2x^{2} - x - 1 = 0,$
$a = 2,b = -1,c = -1,$
$b2-4ac= (-1)^{2} - 4 × 2 × (-1) = 9,$
$x = \frac{1 \pm \sqrt{9}}{2 × 2} = \frac{1 \pm 3}{4},$
$x_{1} = \frac{1 + 3}{4} = 1,$$x_{2} = \frac{1 - 3}{4} = -\frac{1}{2};$
解:項(xiàng)得:$2x^{2} - 7x - 4 = 0,$
$a = 2,b = -7,c = -4,$
$b2-4ac= = (-7)^{2} - 4 × 2 × (-4) = 81,$
$x = \frac{7 \pm \sqrt{81}}{2 × 2} = \frac{7 \pm 9}{4},$
$x_{1} = \frac{7 + 9}{4} = 4,$$x_{2} = \frac{7 - 9}{4} = -\frac{1}{2}。$
$x^{2}-2\sqrt{2}x + 1 $
$= 0$
1
$-2\sqrt{2}$
1
4
$\frac{1\pm\sqrt{13}}{6}$