(x+2)^4+(x+2)^2-12=0

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вопрос

Let's substitute u = x + 2 to simplify the equation:

With this substitution, our equation becomes
u^4 + u^2 - 12 = 0

This is a quadratic equation in u. Let's solve for u using the quadratic formula:

u = [-b ± √(b^2 - 4ac)] / 2a

In this case, a=1, b=1, and c=-12.

Plugging those values into the formula:

u = [-1 ± √(1^2 - 4(1)(-12))] / 2(1
u = [-1 ± √(1 + 48)] /
u = [-1 ± √49] /
u = [-1 ± 7] / 2

Therefore, we have two possible solutions for u
u = (7 - 1) / 2 = 6 / 2 =
u = (-1 - 7) / 2 = -8 / 2 = -4

Now, substitute back x + 2 for u
For u = 3, x + 2 =
x = 1

For u = -4, x + 2 = -
x = -6

Therefore, the solutions to the equation (x+2)^4 + (x+2)^2 - 12 = 0 are x = 1 and x = -6.