(x^2-4)^2+(x^2-6x-16)^2=0

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вопрос

To solve the given equation:

(x^2 - 4)^2 + (x^2 - 6x - 16)^2 = 0

Let's first expand the squares:

(x^4 - 8x^2 + 16) + (x^4 - 12x^3 + 36x^2 - 12x^2 + 72x + 256) = 0

Combine like terms:

2x^4 - 20x^2 - 12x^3 + 72x + 272 = 0

Rearrange the terms to get a simpler form:

2x^4 - 12x^3 - 20x^2 + 72x + 272 = 0

Since the degrees of the terms are all even, this is a quartic equation.

Unfortunately, there is no simple way to solve a quartic equation. You can either try to factorize the equation or use numerical methods to find approximate solutions.

If you have a specific value for x that you would like to substitute in and check the validity of the equation, please provide it.