Question

(х²+2)²+2(х²+х)-8=0 ​

Answer 1

Ответ:ОбъяснениеhОбъяснениеhn

Объяснение:

Answer 2

To solve the quadratic equation $(x^2 + 2)^2 + 2(x^2 + x) - 8 = 0$, we can follow these steps:

1. Expand and simplify the equation:

   $$(x^2 + 2)^2 + 2(x^2 + x) - 8 = 0$$
   $$x^4 + 4x^2 + 4 + 2x^2 + 2x - 8 = 0$$
   $$x^4 + 6x^2 + 2x - 4 = 0$$

2. Substitute $y = x^2$ to transform the quartic equation into a quadratic equation:

   $$y^2 + 6y + 2x - 4 = 0$$

3. Solve this quadratic equation for $y$ using the quadratic formula:

   $$y = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times (-4)}}{2 \times 1}$$
   $$y = \frac{-6 \pm \sqrt{52}}{2}$$
   $$y_1 = -3 + \sqrt{13}, \quad y_2 = -3 - \sqrt{13}$$

4. Substitute $y$ back in terms of $x$:

   $$x^2 = -3 + \sqrt{13}, \quad x^2 = -3 - \sqrt{13}$$

5. Solve for $x$ by taking the square root of both sides:

   $$x = \pm \sqrt{-3 + \sqrt{13}}, \quad x = \pm \sqrt{-3 - \sqrt{13}}$$

These are the four solutions for the given equation.