Question

(x + 2) 4+3 (x+2) 2-4=0​

Answer 1

Объяснение:

4х+8+6х+12-4=0

10х=-16

х=-1,6

Answer 2

It seems like you have an equation that you'd like to solve for the variable "x." The equation you provided is:

(x + 2)^4 + 3(x + 2)^2 - 4 = 0

To solve this equation, we can use a substitution method. Let's make a substitution:

Let y = (x + 2)^2

Now, we can rewrite the equation in terms of y:

y^2 + 3y - 4 = 0

This is a quadratic equation in terms of y. We can factor it:

(y + 4)(y - 1) = 0

Now, we can solve for y:

1. y + 4 = 0 y = -4

2. y - 1 = 0 y = 1

Now that we have two possible values for y, we need to substitute back to find the corresponding values for x:

1. For y = -4: (x + 2)^2 = -4

   Since the square of a real number can't be negative, there are no real solutions in this case.

2. For y = 1: (x + 2)^2 = 1

   Taking the square root of both sides:

   x + 2 = ±1

Now, we have two separate equations to solve:

a. x + 2 = 1 x = 1 - 2 x = -1

b. x + 2 = -1 x = -1 - 2 x = -3

So, the solutions to the original equation are:

x = -1 and x = -3.