532. Решите уравнение: a) 3x² - 7x + 4 = 0; 6) 5x² - 8x + 3 = 0; в) 3x² - 13х + 14 = 0; г) 2y² - 9y + 10 = 0; д) 5y² - 6y + 1 = 0; e) 4x² + x - 33 = 0; ж) y² - 10y - 24 = 0; з) p² + p - 90 = 0.

Решение:

Для квадратного уравнения вида $$ax^2+bx+c=0$$ корни можно найти по формуле: $$x = \frac{-b \pm \sqrt{D}}{2a}$$, где дискриминант $$D = b^2 - 4ac$$.

1. a) $$3x^2 - 7x + 4 = 0$$ $$D = (-7)^2 - 4 \cdot 3 \cdot 4 = 49 - 48 = 1$$ $$x_1 = \frac{7 + \sqrt{1}}{2 \cdot 3} = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}$$ $$x_2 = \frac{7 - \sqrt{1}}{2 \cdot 3} = \frac{7 - 1}{6} = \frac{6}{6} = 1$$
2. б) $$5x^2 - 8x + 3 = 0$$ $$D = (-8)^2 - 4 \cdot 5 \cdot 3 = 64 - 60 = 4$$ $$x_1 = \frac{8 + \sqrt{4}}{2 \cdot 5} = \frac{8 + 2}{10} = \frac{10}{10} = 1$$ $$x_2 = \frac{8 - \sqrt{4}}{2 \cdot 5} = \frac{8 - 2}{10} = \frac{6}{10} = \frac{3}{5}$$
3. в) $$3x^2 - 13x + 14 = 0$$ $$D = (-13)^2 - 4 \cdot 3 \cdot 14 = 169 - 168 = 1$$ $$x_1 = \frac{13 + \sqrt{1}}{2 \cdot 3} = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3}$$ $$x_2 = \frac{13 - \sqrt{1}}{2 \cdot 3} = \frac{13 - 1}{6} = \frac{12}{6} = 2$$
4. г) $$2y^2 - 9y + 10 = 0$$ $$D = (-9)^2 - 4 \cdot 2 \cdot 10 = 81 - 80 = 1$$ $$y_1 = \frac{9 + \sqrt{1}}{2 \cdot 2} = \frac{9 + 1}{4} = \frac{10}{4} = \frac{5}{2}$$ $$y_2 = \frac{9 - \sqrt{1}}{2 \cdot 2} = \frac{9 - 1}{4} = \frac{8}{4} = 2$$
5. д) $$5y^2 - 6y + 1 = 0$$ $$D = (-6)^2 - 4 \cdot 5 \cdot 1 = 36 - 20 = 16$$ $$y_1 = \frac{6 + \sqrt{16}}{2 \cdot 5} = \frac{6 + 4}{10} = \frac{10}{10} = 1$$ $$y_2 = \frac{6 - \sqrt{16}}{2 \cdot 5} = \frac{6 - 4}{10} = \frac{2}{10} = \frac{1}{5}$$
6. e) $$4x^2 + x - 33 = 0$$ $$D = (1)^2 - 4 \cdot 4 \cdot (-33) = 1 + 528 = 529$$ $$x_1 = \frac{-1 + \sqrt{529}}{2 \cdot 4} = \frac{-1 + 23}{8} = \frac{22}{8} = \frac{11}{4}$$ $$x_2 = \frac{-1 - \sqrt{529}}{2 \cdot 4} = \frac{-1 - 23}{8} = \frac{-24}{8} = -3$$
7. ж) $$y^2 - 10y - 24 = 0$$ $$D = (-10)^2 - 4 \cdot 1 \cdot (-24) = 100 + 96 = 196$$ $$y_1 = \frac{10 + \sqrt{196}}{2 \cdot 1} = \frac{10 + 14}{2} = \frac{24}{2} = 12$$ $$y_2 = \frac{10 - \sqrt{196}}{2 \cdot 1} = \frac{10 - 14}{2} = \frac{-4}{2} = -2$$
8. з) $$p^2 + p - 90 = 0$$ $$D = (1)^2 - 4 \cdot 1 \cdot (-90) = 1 + 360 = 361$$ $$p_1 = \frac{-1 + \sqrt{361}}{2 \cdot 1} = \frac{-1 + 19}{2} = \frac{18}{2} = 9$$ $$p_2 = \frac{-1 - \sqrt{361}}{2 \cdot 1} = \frac{-1 - 19}{2} = \frac{-20}{2} = -10$$

Ответ:

1. a) x₁ = 4/3, x₂ = 1
2. б) x₁ = 1, x₂ = 3/5
3. в) x₁ = 7/3, x₂ = 2
4. г) y₁ = 5/2, y₂ = 2
5. д) y₁ = 1, y₂ = 1/5
6. e) x₁ = 11/4, x₂ = -3
7. ж) y₁ = 12, y₂ = -2
8. з) p₁ = 9, p₂ = -10