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電子課本網(wǎng) 第10頁

第10頁

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1
$ -\sqrt{5}-2 $
A
解:將方程化為一般形式:$2x^2 - 7x - 4 = 0,$其中$a = 2,$$b = -7,$$c = -4。$
判別式$\Delta = b^2 - 4ac = (-7)^2 - 4 \times 2 \times (-4) = 49 + 32 = 81。$
由求根公式$x = \frac{-b \pm \sqrt{\Delta}}{2a},$得:
$x = \frac{7 \pm \sqrt{81}}{2 \times 2} = \frac{7 \pm 9}{4},$
即$x_1 = \frac{7 + 9}{4} = 4,$$x_2 = \frac{7 - 9}{4} = -\frac{1}{2}。$
解:方程$-2x^2 + 2x - 1 = 0$中,$a = -2,$$b = 2,$$c = -1。$
判別式$\Delta = b^2 - 4ac = 2^2 - 4 \times (-2) \times (-1) = 4 - 8 = -4。$
因為$\Delta < 0,$所以方程無實數(shù)根。
解:將方程化為一般形式:$0.3y^2 + y - 0.8 = 0,$其中$a = 0.3,$$b = 1,$$c = -0.8。$
判別式$\Delta = b^2 - 4ac = 1^2 - 4 \times 0.3 \times (-0.8) = 1 + 0.96 = 1.96。$
由求根公式$y = \frac{-b \pm \sqrt{\Delta}}{2a},$得:
$y = \frac{-1 \pm \sqrt{1.96}}{2 \times 0.3} = \frac{-1 \pm 1.4}{0.6},$
即$y_1 = \frac{-1 + 1.4}{0.6} = \frac{0.4}{0.6} = \frac{2}{3},$$y_2 = \frac{-1 - 1.4}{0.6} = \frac{-2.4}{0.6} = -4。$
解:將方程化為一般形式:$-\frac{2}{3}x^2 + \frac{1}{2}x + 1 = 0,$其中$a = -\frac{2}{3},$$b = \frac{1}{2},$$c = 1。$
判別式$\Delta = b^2 - 4ac = \left(\frac{1}{2}\right)^2 - 4 \times \left(-\frac{2}{3}\right) \times 1 = \frac{1}{4} + \frac{8}{3} = \frac{3 + 32}{12} = \frac{35}{12}。$
由求根公式$x = \frac{-b \pm \sqrt{\Delta}}{2a},$得:
$x = \frac{-\frac{1}{2} \pm \sqrt{\frac{35}{12}}}{2 \times \left(-\frac{2}{3}\right)} = \frac{-\frac{1}{2} \pm \frac{\sqrt{105}}{6}}{-\frac{4}{3}} = \frac{-\frac{3}{6} \pm \frac{\sqrt{105}}{6}}{-\frac{4}{3}} = \frac{-3 \pm \sqrt{105}}{6} \times \left(-\frac{3}{4}\right) = \frac{3 \mp \sqrt{105}}{8},$
即$x_1 = \frac{3 + \sqrt{105}}{8},$$x_2 = \frac{3 - \sqrt{105}}{8}。$
解:$3x^2 + 5(2x + 1) = 0$
展開得:$3x^2 + 10x + 5 = 0$
其中$a = 3,$$b = 10,$$c = 5$
判別式$\Delta = b^2 - 4ac = 10^2 - 4 \times 3 \times 5 = 100 - 60 = 40$
由求根公式得:$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-10 \pm \sqrt{40}}{2 \times 3} = \frac{-10 \pm 2\sqrt{10}}{6} = \frac{-5 \pm \sqrt{10}}{3}$
所以$x_1 = \frac{-5 + \sqrt{10}}{3},$$x_2 = \frac{-5 - \sqrt{10}}{3}$
解:$(x - 3)^2 + 2x(x - 3) = 0$
提取公因式$(x - 3)$得:$(x - 3)[(x - 3) + 2x] = 0$
化簡括號內(nèi)得:$(x - 3)(3x - 3) = 0$
即$x - 3 = 0$或$3x - 3 = 0$
解得$x_1 = 3,$$x_2 = 1$
解:設$x^2 = y,$則原方程可化為$y^2 - y - 6 = 0。$解得$y_1 = 3,$$y_2 = -2$(不合題意,舍去)。由$x^2 = 3$可得解是:$x_1 = \sqrt{3},$$x_2 = -\sqrt{3},$故方程$x^4 - x^2 - 6 = 0$的解是$x_1 = \sqrt{3},$$x_2 = -\sqrt{3}。$
解:設$ x^2 + x = y ,$則原方程為$ y(y - 3) = -2 ,$整理得$ y^2 - 3y + 2 = 0 ,$即$(y - 1)(y - 2) = 0,$解得$ y_1 = 1 ,$$ y_2 = 2 。$
當$ x^2 + x = 1 $時,$ x^2 + x - 1 = 0 ,$根據(jù)求根公式$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $(其中$ a = 1 ,$$ b = 1 ,$$ c = -1 $),解得$ x_1 = \frac{-1 + \sqrt{5}}{2} ,$$ x_2 = \frac{-1 - \sqrt{5}}{2} 。$
當$ x^2 + x = 2 $時,$ x^2 + x - 2 = 0 ,$即$(x + 2)(x - 1) = 0,$解得$ x_3 = 1 ,$$ x_4 = -2 。$
綜上,$ x $的值為$ x_1 = \frac{-1 + \sqrt{5}}{2} ,$$ x_2 = \frac{-1 - \sqrt{5}}{2} ,$$ x_3 = 1 ,$$ x_4 = -2 。$