Question

2(2x−8)2−8(2x−8)+8=0

Answer 1

Ответ:

там, если что равно 0

Объяснение:

пример не стала переписыватт

Answer 2

To solve the equation, let's first simplify it step by step:

1. Distribute the square term (2x - 8)²:

(2x - 8)² = (2x - 8)(2x - 8) = 4x² - 16x - 16x + 64 = 4x² - 32x + 64

2. Distribute the -8 to both terms in the parentheses:

-8(2x - 8) = -8 * 2x + (-8) * (-8) = -16x + 64

Now, the equation becomes:

2(4x² - 32x + 64) - 16x + 64 + 8 = 0

Next, simplify further:

1. Distribute the 2 to the terms inside the parentheses:

2(4x²) = 8x² 2(-32x) = -64x 2(64) = 128

So the equation becomes:

8x² - 64x + 128 - 16x + 64 + 8 = 0

Combine like terms:

8x² - 80x + 200 = 0

Now we have a quadratic equation in the form of ax² + bx + c = 0, where a = 8, b = -80, and c = 200.

To solve the quadratic equation, you can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

Plug in the values:

x = (80 ± √((-80)² - 4 * 8 * 200)) / 2 * 8

Calculate inside the square root:

x = (80 ± √(6400 - 6400)) / 16

Since the value inside the square root is zero, we get:

x = (80 ± 0) / 16

There are two solutions:

1. x = (80 + 0) / 16 = 80 / 16 = 5
2. x = (80 - 0) / 16 = 80 / 16 = 5

Both solutions are the same, so the equation has a repeated root:

x = 5