Question

Уравнение (х^2-8)^2+4(х^2-8)-5=0​

Answer 1

Ответ:

Ответ в приложении, решали через замену и дискриминант.

Answer 2

To solve the equation $(x^2-8)^2 + 4(x^2-8) - 5 = 0$, we can use a substitution to simplify the expression and then solve for the variable. Let's make a substitution: let $y = x^2 - 8$.

Now, the equation becomes $y^2 + 4y - 5 = 0$.

This is a quadratic equation in $y$. To solve it, we can use the quadratic formula:

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

where $a = 1$, $b = 4$, and $c = -5$.

$y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot -5}}{2 \cdot 1}$

$y = \frac{-4 \pm \sqrt{16 + 20}}{2}$

$y = \frac{-4 \pm \sqrt{36}}{2}$

$y = \frac{-4 \pm 6}{2}$

Now we have two possibilities for $y$:

1. $y = \frac{-4 + 6}{2} = 1$
2. $y = \frac{-4 - 6}{2} = -5$

Now that we have the possible values for $y$, we can revert back to the original variable $x$ using the substitution we made earlier:

1. $y = x^2 - 8 = 1$ Add 8 to both sides: $x^2 = 9$ Take the square root: $x = \pm 3$

2. $y = x^2 - 8 = -5$ Add 8 to both sides: $x^2 = 3$ Take the square root: $x = \pm \sqrt{3}$

So, the solutions for the original equation are $x = 3$, $x = -3$, $x = \sqrt{3}$, and $x = -\sqrt{3}$.