Задание 46. Решите квадратные уравнения. 7. Важно знать: формула корней квадратного уравнения ах²+bx+c = 0, где а≠ 0

1) \(x^2 + 5x - 14 = 0\)

\(a = 1, b = 5, c = -14\)

\(D = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot (-14) = 25 + 56 = 81\)

\(x_{1,2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{81}}{2 \cdot 1} = \frac{-5 \pm 9}{2}\)

\(x_1 = \frac{-5 + 9}{2} = \frac{4}{2} = 2\)

\(x_2 = \frac{-5 - 9}{2} = \frac{-14}{2} = -7\)

Ответ: \(x_1 = 2, x_2 = -7\)

2) \(12x^2 + 7x + 1 = 0\)

\(a = 12, b = 7, c = 1\)

\(D = b^2 - 4ac = 7^2 - 4 \cdot 12 \cdot 1 = 49 - 48 = 1\)

\(x_{1,2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{1}}{2 \cdot 12} = \frac{-7 \pm 1}{24}\)

\(x_1 = \frac{-7 + 1}{24} = \frac{-6}{24} = -\frac{1}{4} = -0.25\)

\(x_2 = \frac{-7 - 1}{24} = \frac{-8}{24} = -\frac{1}{3}\)

Ответ: \(x_1 = -0.25, x_2 = -\frac{1}{3}\)

3) \(6x^2 - x - 1 = 0\)

\(a = 6, b = -1, c = -1\)

\(D = ?\)

4) \(4x^2 + 4x + 1 = 0\)

\(a = 4, b = 4, c = 1\)

\(D = ?\)

6) \(-x^2 - 3x + 4 = 0\) или \(x^2 + 3x - 4 = 0\)

\(a = 1, b = 3, c = -4\)

\(D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-4) = 9 + 16 = 25\)

\(x_{1,2} = \frac{-b \pm \sqrt{D}}{2a} = \frac{-3 \pm \sqrt{25}}{2 \cdot 1} = \frac{-3 \pm 5}{2}\)

\(x_1 = \frac{-3 + 5}{2} = \frac{2}{2} = 1\)

\(x_2 = \frac{-3 - 5}{2} = \frac{-8}{2} = -4\)

Ответ: \(x_1 = 1, x_2 = -4\)

7) \(2x^2 - 2x + 5 = 0\)

\(a = 2, b = -2, c = 5\)

\(D = b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot 5 = 4 - 40 = -36\)

Дискриминант меньше нуля: нет корней.

8) \(x^2 + 7x - 8 = 0\)

\(a = 1, b = 7, c = -8\)

\(D = ?\)

9) \(6x^2 + 7x + 2 = 0\)

\(a = 6, b = 7, c = 2\)

\(D = ?\)