Question

(х-2)^4-(х-2)^2-6=0​

Answer 1

Это биквадратное уравнение.

(х-2)^2=t

t^2-t-6=0

D=1+24=25

t1=(1+5)/2=3

t2=(1-5)2=-2

2) (x-2)^2 не может быть отрицательным, поэтому 2ой корень не подходит

3) (х-2)^2=3

х^2-4х+1=0

D1=4-1=3

x1=2+√3

x2=2-√3
Ответ: 2+√3, 2-√3

Answer 2

To solve the equation (х-2)^4 - (х-2)^2 - 6 = 0, we can use a substitution to make the calculation easier. Let's substitute a new variable, u, for (х-2)^2. This means u = (х-2)^2.

Now, the equation becomes:

u^2 - u - 6 = 0

This is a quadratic equation in terms of u. To solve it, we can factor it or use the quadratic formula. Factoring, we get:

(u + 2)(u - 3) = 0

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

1. u + 2 = 0 u = -2

2. u - 3 = 0 u = 3

Remember that u = (х-2)^2, so we have:

1. (х-2)^2 = -2 There are no real solutions for this case, as the square of a real number cannot be negative.

2. (х-2)^2 = 3

Now, take the square root of both sides:

√((х-2)^2) = ±√3

Simplify:

х - 2 = ±√3

Now, solve for х:

1. х - 2 = √3 х = 2 + √3

2. х - 2 = -√3 х = 2 - √3

So, the solutions to the equation are х = 2 + √3 and х = 2 - √3.