Question

(х-6)(х+6)+х(8-2х)+20=0
(3х-1)(3х+1)-(3-х)²+(2х+1)²=13
знайти корінь рівняння

Answer 1

Ответ:

1. x1 =4;

• x2=4;

2. x1=1;

x2=-11/6

Answer 2

Ответ:

1) (x - 6)(x + 6) + x(8 - 2x) + 20 = 0

x² - 36 + 8x - 2x² + 20 = 0

-x² + 8x - 16 = 0

x² - 8x + 16 = 0

D = (-8)² - 4 × 1 × 16 = 64 - 64 = 0

x¹ =

$\frac{8}{2 \times 1} = \frac{8}{2} = 4$

2) (3x - 1)(3x + 1) - (3 - x)² + (2x + 1)² = 13

9x² - 1 - (9 - 9x + x²) + 4x² + 4x + 1 = 13

9x² - 1 - 9 + 9x - x² + 4x² + 4x + 1 = 13

12x² + 13x - 22 = 0

D = 13² - 4 × 12 × (-22) = 169 + 1056 = 1225

x¹ =

$\frac{ - 13 + \sqrt{1225} }{2 \times 12} = \frac{ - 13 + 35}{24} = \frac{22}{24} = \frac{11}{12}$

x² =

$\frac{ - 13 - \sqrt{1225} }{2 \times 12} = \frac{ - 13 - 35}{24} = \frac{ - 48}{24} = - 2$