Будет равно : 0,6an^39 n^91
<em>Уравнение имеет вид ax² + bx + c = 0, где a, b, c -- коэффициенты.</em>
<em>Дискриминант находится по формуле: D = b² - 4ac</em>
<em>Корни уравнения находятся по формулам: </em>
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<em>Остаётся подставить в формулы и вычислить необходимое.</em>
1. 3x² + 2x - 5 = 0
a = 3, b = 2, c = -5
D = 2² - 4 * 3 * (-5) = 4 + 60 = 64
![x_1=\frac{-2-\sqrt{64}}{2\cdot3}=\frac{-2-8}{6}=-\frac{5}{3}\\ \\ x_2=\frac{-2+\sqrt{64}}{2\cdot3}=\frac{-2+8}{6}=1](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B-2-%5Csqrt%7B64%7D%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B-2-8%7D%7B6%7D%3D-%5Cfrac%7B5%7D%7B3%7D%5C%5C%20%5C%5C%20x_2%3D%5Cfrac%7B-2%2B%5Csqrt%7B64%7D%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B-2%2B8%7D%7B6%7D%3D1)
2. 9x² - 6x + 1 = 0
a = 9, b = -6, c = 1
D = (-6)² - 4 * 9 * 1 = 36 - 36 = 0
![x_1=\frac{6-\sqrt{0}}{2\cdot9}=\frac{6-0}{18}=\frac{1}{3}\\ \\ x_2=\frac{6+\sqrt{0}}{2\cdot9}=\frac{6+0}{18}=\frac{1}{3}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B6-%5Csqrt%7B0%7D%7D%7B2%5Ccdot9%7D%3D%5Cfrac%7B6-0%7D%7B18%7D%3D%5Cfrac%7B1%7D%7B3%7D%5C%5C%20%5C%5C%20x_2%3D%5Cfrac%7B6%2B%5Csqrt%7B0%7D%7D%7B2%5Ccdot9%7D%3D%5Cfrac%7B6%2B0%7D%7B18%7D%3D%5Cfrac%7B1%7D%7B3%7D)
3. 5x² - x - 4 = 0
a = 5, b = -1, c = -4
D = (-1)² - 4 * 5 * (-4) = 1 + 80 = 81
![x_1=\frac{1-\sqrt{81}}{2\cdot5}=\frac{1-9}{10}=-\frac{4}{5}\\ \\ x_2=\frac{1+\sqrt{81}}{2\cdot5}=\frac{1+9}{10}=1](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B1-%5Csqrt%7B81%7D%7D%7B2%5Ccdot5%7D%3D%5Cfrac%7B1-9%7D%7B10%7D%3D-%5Cfrac%7B4%7D%7B5%7D%5C%5C%20%5C%5C%20x_2%3D%5Cfrac%7B1%2B%5Csqrt%7B81%7D%7D%7B2%5Ccdot5%7D%3D%5Cfrac%7B1%2B9%7D%7B10%7D%3D1)
4. 3x² + 2x - 1 = 0
a = 3, b = 2, c = -1
D = 2² - 4 * 3 * (-1) = 4 + 12 = 16
![x_1=\frac{-2-\sqrt{16}}{2\cdot3}=\frac{-2-4}{6}=-1 \\ \\ x_2=\frac{-2+\sqrt{16}}{2\cdot3}=\frac{-2+4}{6}=\frac{1}{3}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B-2-%5Csqrt%7B16%7D%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B-2-4%7D%7B6%7D%3D-1%20%5C%5C%20%5C%5C%20x_2%3D%5Cfrac%7B-2%2B%5Csqrt%7B16%7D%7D%7B2%5Ccdot3%7D%3D%5Cfrac%7B-2%2B4%7D%7B6%7D%3D%5Cfrac%7B1%7D%7B3%7D)
5. <em>Повторяет пункт 3.</em>