Question

|x^2-16|+|x+2|=-x^2-x+14

Answer 1

Ответ:  [ -4 ; -2]

Пошаговое объяснение:

Answer 2

To solve this equation, we need to consider several cases depending on the sign of the expressions inside the absolute value bars.

Case 1: x^2 - 16 is non-negative and x + 2 is non-negative In this case, the equation simplifies to: x^2 - 16 + x + 2 = -x^2 - x + 14 Simplifying further, we get: 2x^2 + 2x - 12 = 0 Dividing by 2, we get: x^2 + x - 6 = 0 Factoring, we get: (x + 3)(x - 2) = 0 So, x = -3 or x = 2.

Case 2: x^2 - 16 is non-negative and x + 2 is negative In this case, the equation simplifies to: x^2 - 16 - x - 2 = -x^2 - x + 14 Simplifying further, we get: 2x^2 + 2x - 6 = 0 Dividing by 2, we get: x^2 + x - 3 = 0 Factoring, we get: (x + 3)(x - 1) = 0 So, x = -3 or x = 1.

Case 3: x^2 - 16 is negative and x + 2 is non-negative In this case, the equation simplifies to: 16 - x^2 + x + 2 = -x^2 - x + 14 Simplifying further, we get: 2x^2 + 2x - 4 = 0 Dividing by 2, we get: x^2 + x - 2 = 0 Factoring, we get: (x + 2)(x - 1) = 0 So, x = -2 or x = 1.

Case 4: x^2 - 16 is negative and x + 2 is negative In this case, the equation simplifies to: 16 - x^2 - x - 2 = -x^2 - x + 14 Simplifying further, we get: x^2 - 15 = 0 Taking the square root, we get: x = ±√15 But since x^2 - 16 is negative in this case, we must discard these solutions.

Therefore, the only solutions that satisfy the original equation are x = -3 and x = 1.