28.17 a) x²+x+1=0; 6) x² + 5x + 2¼ = 0;

а) 1/3 x² + x + 1/4 = 0

Умножим на 12 чтобы избавиться от дробей:

4x² + 12x + 3 = 0

a = 4, b = 12, c = 3

$$D = b^2 - 4ac = 12^2 - 4 \cdot 4 \cdot 3 = 144 - 48 = 96$$

$$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-12 + \sqrt{96}}{2 \cdot 4} = \frac{-12 + 4\sqrt{6}}{8} = \frac{-3 + \sqrt{6}}{2}$$

$$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-12 - \sqrt{96}}{2 \cdot 4} = \frac{-12 - 4\sqrt{6}}{8} = \frac{-3 - \sqrt{6}}{2}$$

б) x² + 5x + 2¼ = 0

x² + 5x + 9/4 = 0

a = 1, b = 5, c = 9/4

$$D = b^2 - 4ac = 5^2 - 4 \cdot 1 \cdot \frac{9}{4} = 25 - 9 = 16$$

$$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-5 + \sqrt{16}}{2 \cdot 1} = \frac{-5 + 4}{2} = \frac{-1}{2}$$

$$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-5 - \sqrt{16}}{2 \cdot 1} = \frac{-5 - 4}{2} = \frac{-9}{2}$$

Ответ: a) x₁ = (-3 + √6)/2, x₂ = (-3 - √6)/2; б) x₁ = -1/2, x₂ = -9/2