Question

3. (4x² – 8)2 + 11(4x² – 8) + 28=0.​

Answer 1

Ответ:

Я так решила, надеюсь что правильно

Answer 2

To solve the quadratic equation (4x² - 8)² + 11(4x² - 8) + 28 = 0, we can use a substitution to simplify the equation. Let's set y = 4x² - 8, so our equation becomes:

y² + 11y + 28 = 0

Now, we have a quadratic equation in terms of y. To solve it, we can factor it or use the quadratic formula. In this case, let's factor it:

(y + 4)(y + 7) = 0

Now, we have two possible values for y:

1. y + 4 = 0 y = -4

2. y + 7 = 0 y = -7

Now, we need to revert back to the original variable, x:

1. For y = -4: 4x² - 8 = -4 4x² = 4 x² = 1 x = ±√1 x = ±1

2. For y = -7: 4x² - 8 = -7 4x² = 1 x² = 1/4 x = ±√(1/4) x = ±(1/2)

So, the solutions to the original equation (4x² - 8)² + 11(4x² - 8) + 28 = 0 are: x = 1, -1, 1/2, -1/2