Question

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

Answer 1

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

(x-2)^2 = t

t^2-t-12=0

D=1+48=49=7^2

t=(1+-7)/2

t1=4
t2=-3

obratnaia zamena

(x-2)^2=4

(x-2)^2=-3

x1=4

x2=0

Answer 2

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

Substituting u into the equation, we have:

u^2 - u - 12 = 0.

Now we have a quadratic equation in terms of u. To solve for u, we can factorize the quadratic equation:

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

Setting each factor to zero, we get:

u - 4 = 0 --> u = 4, u + 3 = 0 --> u = -3.

Now, we substitute back u = (x - 2)^2 into these solutions:

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

Taking the square root of both sides, we have two possibilities:

x - 2 = 2 --> x = 4, x - 2 = -2 --> x = 0.

For u = -3: (x - 2)^2 = -3.

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

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