Question

(X+3)^2+(X-4)^2=2(4-x(x+4)

Answer 1

Ответ:

Объяснение:

ответ и поЯснение на фото

Answer 2

To solve the equation $(x+3)^2 + (x-4)^2 = 2(4-x(x+4))$, you can follow these steps:

1. Expand both sides of the equation:

$(x+3)^2 + (x-4)^2 = 2(4-x(x+4))$

Expanding the left side:

$(x^2 + 6x + 9) + (x^2 - 8x + 16) = 2(4 - x^2 - 4x)$

2. Combine like terms on both sides:

$(2x^2 - 2x + 25) = 2(4 - x^2 - 4x)$

3. Distribute the 2 on the right side of the equation:

$2x^2 - 2x + 25 = 8 - 2x^2 - 8x$

4. Move all terms to one side to set the equation to zero:

$2x^2 - 2x + 25 - 8 + 2x^2 + 8x = 0$

5. Combine like terms:

$4x^2 + 25 - 8 = 0$

$4x^2 + 17 = 0$

6. Subtract 17 from both sides:

$4x^2 = -17$

7. Divide both sides by 4:

$x^2 = -\frac{17}{4}$

8. Take the square root of both sides, remembering to consider both the positive and negative square roots:

$x = \pm \sqrt{-\frac{17}{4}}$

Since the right side of the equation contains a negative number under the square root, this equation does not have real solutions. In the real number system, there are no real values of x that satisfy the equation. The solutions are complex numbers:

$x = \pm \frac{i}{2}\sqrt{17}$

So, the solutions are $x = \frac{i}{2}\sqrt{17}$ and $x = -\frac{i}{2}\sqrt{17}$, where $i$ is the imaginary unit.