Question

Надите корьни уровнения 1)(2x-3)^2=11x-19
2)(x-2)^2=-2x+31

Answer 1

Ответ:

(2x-3)² = 11x - 19

4x²-12x + 9 = 11x - 19

4x² - 12x - 11x + 9 + 19 = 0

4x² - 23x + 28 = 0

a=4, b=-23, c=28

D = b² - 4ac

D = (-23)² - 4×4×28= 529 - 448

D = 81

√D=9

x₁ =

$\frac{23 + 9}{2 \times 4} = \frac{32}{8} = 4$

x₂ =

$\frac{23 - 9}{2 \times 4} = \frac{14}{8} =1.75$

Ответ: x ∈ (4; 1,75)

2)

(x-2)² = -2x + 31

x²-4x + 4 = -2x + 31

x² - 4x +2x +4 - 31 = 0

x² - 2x - 27 = 0

a=1, b=-2, c= -27

D= b² - 4ac

D=(-2)² - 4 × 1 × (-27) = 4 + 108

D = 112

√D = √112

x₁ =

$\frac{ 2 + \sqrt{112} }{2 \times 1} = \frac{ 2 + \sqrt{7 \times 16} }{2} = \frac{2 + 4 \sqrt{7} }{2} = 1 + 2 \sqrt{7}$

x₂ =

$\frac{2 - \sqrt{112} }{2 \times 1} = \frac{2 - \sqrt{7 \times 16} }{2} = \frac{2 - 4 \sqrt{7} }{2} = 1 - 2 \sqrt{7}$

$Ответ: x ∈ (1 + 2 \sqrt{7} ;1 - 2 \sqrt{7} )$