Question

(х^2 – 5)^2–2(х^2 – 5) − 8 = 0.

Answer 1

Решение:
$( x^{2} - 5 )^{2} -2*(x^{2} - 5)- 8 = 0$
Пусть  $x^{2} - 5 = t$, получим
$t^{2} -2t - 8 = 0$
$D = 4 + 32 = 36 = 6^{2}$
$t_{1} = \frac{2 + 6}{2} = 4$
$t_{2} = \frac{2 - 6}{2} = -2$
Получили, что 
$x^{2} - 5 = 4$ или  $x^{2} - 5 = - 2$
1)  $x^{2} - 5 = 4$
$x^{2} = 9$
$x = 3$или  $x = - 3$
2)  $x^{2} - 5 = - 2$
$x^{2} = 5 - 2$
$x^{2} = 3$
$x = \sqrt{3}$ или  $x = - \sqrt{3}$
Ответ: -3; -√3; √3; 3.
Проверка:
Если х = ± 3, то  $( 9 - 5 )^{2} -2*(9 - 5) - 8 = 0; 16 - 8 - 8 = 0$
 - верно.
Если х = ± √3, то  $( 3 - 5 )^{2} -2*(3 - 5) - 8 =0;4+4-8 = 0$
 - верно.

Answer 2

To solve the quadratic equation $(x^2 - 5)^2 - 2(x^2 - 5) - 8 = 0$, we can use substitution to simplify it. Let's substitute $u = x^2 - 5$. Now the equation becomes:

$u^2 - 2u - 8 = 0$

This is a quadratic equation in terms of $u$. To solve for $u$, we can use the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -2$, and $c = -8$. Plugging in these values:

$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}$

$u = \frac{2 \pm \sqrt{4 + 32}}{2}$

$u = \frac{2 \pm \sqrt{36}}{2}$

$u = \frac{2 \pm 6}{2}$

Now we have two possible values for $u$:

1. $u = \frac{2 + 6}{2} = \frac{8}{2} = 4$

2. $u = \frac{2 - 6}{2} = \frac{-4}{2} = -2$

Now, remember that we initially substituted $u = x^2 - 5$. So we need to set these two values equal to $x^2 - 5$ and solve for $x$:

1. $x^2 - 5 = 4$

$x^2 = 4 + 5$

$x^2 = 9$

$x = \pm \sqrt{9}$

$x = \pm 3$

2. $x^2 - 5 = -2$

$x^2 = -2 + 5$

$x^2 = 3$

$x = \pm \sqrt{3}$

So, the solutions to the given quadratic equation are $x = 3, -3, \sqrt{3}, -\sqrt{3}$.