Question

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

Answer 1

(x+2)^4+(x+2)^2-12=0
пусть (x+2)^2 = t , t≥0
t²+t-12=0
по теореме Виета
t=3
t=-4(пост. кор. по ограничению)
(x+2)^2=3
x²+4x+4=3
x²+4x+1=0
D₁=4-1=3
x=-2 ± √3

Answer 2

To solve the equation (x+2)^4 + (x+2)^2 - 12 = 0, we can use a substitution to simplify the equation. Let's introduce a new variable u = (x+2)^2.

Substituting u in place of (x+2)^2, the equation becomes:

u^2 + u - 12 = 0

Now we have a quadratic equation in terms of u. We can solve it by factoring or using the quadratic formula. Let's factor it:

(u + 4)(u - 3) = 0

Now, we set each factor equal to zero and solve for u:

u + 4 = 0 or u - 3 = 0

Solving these equations, we find:

u = -4 or u = 3

Recall that u = (x+2)^2, so we substitute these values back into the equation:

For u = -4: (x+2)^2 = -4

Since the square of a real number cannot be negative, there are no solutions for this case.

For u = 3: (x+2)^2 = 3

Taking the square root of both sides, we get:

x + 2 = ±√3

Solving for x, we have two cases:

Case 1: x + 2 = √3 x = √3 - 2

Case 2: x + 2 = -√3 x = -√3 - 2

So the solutions to the equation are: x = √3 - 2 x = -√3 - 2