Question

(2-x)(x-4)=-8 (x-5)(x-4)=20 (8-x)(x-2)=10x (1-x)(9-x)=-10x Квадратное уравнение

Answer 1

$1)\; \; (2-x)(x-4)=-8\\\\2x-8-x^2+4x=-8\; \; ,\; \; x^2-6x=0\; \; ,\; \; x(x-6)=0\\\\x_1=0\; ,\; \; x_2=6\\\\2)\; \; (x-5)(x-4)=20\\\\x^2-9x+20=20\; \; ,\; \; x^2-9x=0\; \; ,\; \; x(x-9)=0\\\\x_1=0\; \; ,\; \; x_2=9\\\\3)\; \; (8-x)(x-2)=10x\\\\10x-16-x^2=10x\; \; ,\; \; x^2+16=0\; \; ,\; \; x^2=-16$

Answer 2

1) (2 - x)(x - 4) = - 8

2x - 8 - x² + 4x + 8 = 0

- x² + 6x = 0

x² - 6x = 0

x(x - 6) = 0

или     x₁ = 0

или     x - 6 = 0  и тогда   x₂ = 6

Ответ : 0 ; 6

2) (x - 5)(x - 4) = 20

x² - 4x - 5x + 20 - 20 = 0

x² - 9x = 0

x(x - 9) = 0

или    x₁ =0

или    x - 9 = 0  и тогда    x₂ = 9

Ответ : 0 ; 9

3)(8 - x)(x - 2) = 10x

8x - 16 - x² + 2x - 10x = 0

- x² + 10x - 16 - 10x = 0

- x² - 16 = 0

x² + 16 = 0 - решений нет

4) (1 - x)(9 - x) = - 10x

9 - x - 9x + x² + 10x = 0

x² - 10x + 10x + 9 = 0

x² + 9 = 0 - решений нет

Answer 3

To solve a quadratic equation, we need to rearrange the terms so that the equation is in the form of ax^2 + bx + c = 0, where a, b, and c are constants.

Let's take the first equation, (2-x)(x-4)=-8, and simplify it:

(2-x)(x-4) = -8 2x - 8 - x^2 + 4x = -8 // multiply out the brackets and rearrange -x^2 + 6x - 8 = -8 -x^2 + 6x = 0 x(-x + 6) = 0 x = 0 or x = 6

Similarly, let's simplify the second equation, (x-5)(x-4)=20:

(x-5)(x-4) = 20 x^2 - 9x + 20 = 0 // expand the brackets and simplify (x-4)(x-5) = 0 x = 4 or x = 5

Moving on to the third equation, (8-x)(x-2)=10x:

(8-x)(x-2) = 10x 8x - 2x^2 - 16 + 2x = 10x // expand the brackets and simplify -2x^2 + 10x - 16 = 0 x^2 - 5x + 8 = 0 // divide everything by -2 and rearrange (x-4)(x-2) = 0 x = 4 or x = 2

Finally, let's simplify the fourth equation, (1-x)(9-x)=-10x:

(1-x)(9-x) = -10x 9x - x^2 - x + x^2 = -10x // expand the brackets and simplify 9x - x = -10x 20x = 0 x = 0

Therefore, the solutions to the given set of equations are: x = 0, 2, 4, 5, or 6.