Question

(x^2-2)^2-2(x^2-2)=0​

Answer 1

(x² - 2)² - 2(x² - 2) = 0

Раскроем скобки.

(x²)² - 2 · 2x² + 2² - 2x² + 2 · 2 = 0

Упростим.

x⁴ - 4x² + 4 - 2x² + 4 = 0

Складываем подобные слагаемые.

x⁴ - 6x² + 8 = 0

Решаем как уравнение. Используем замену переменной.

x⁴ - 6x² + 8 = 0

x² = y

y² - 6y + 8 = 0

D = (-6)² - 4 · 1 · 8 = 36 - 32 = 4 = 2²

y1 = (6 - √(2²)) / (2 · 1) = (6 - 2) / 2 = 4 / 2 = 2

y2 = (6 + √(2²)) / (2 · 1) = (6 + 2) / 2 = 8 / 2 = 4

x² = 2; x1 = √2; x2 = -√2;

x² = 4; x3 = 2; x4 = -2.

Ответ: x1 = √2; x2 = -√2; x3 = 2; x4 = -2.

Answer 2

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

Let $y = x^2 - 2$. Substituting this into the equation, we get:

$y^2 - 2y = 0$

Now, factor out $y$ from the left side:

$y(y - 2) = 0$

This equation has two solutions for $y$:

1. $y = 0$
2. $y - 2 = 0$, which implies $y = 2$

Now, remember that $y = x^2 - 2$. So, we substitute the values of $y$ back into this equation to find the corresponding values of $x$:

1. $y = 0$: $x^2 - 2 = 0$ $x^2 = 2$ $x = \pm \sqrt{2}$

2. $y = 2$: $x^2 - 2 = 2$ $x^2 = 4$ $x = \pm 2$

So, the solutions to the original equation are $x = \pm \sqrt{2}$ and $x = \pm 2$.