Question

(x+2)^4-5(x+2)^2+4=0

Answer 1

Ответ:

-4; -3; -1; 0.

Объяснение:

(x+2)^4-5(x+2)^2+4=0

Пусть (х+2)² = t, t ≥ 0, тогда

t² - 5t + 4 = 0

D = 25 - 16 = 9

t1 = (5+3)/2 = 4;

t2 = (5-3)/2 = 1.

Получили, что

(х+2)² = 1 или (х+2)² = 4

1) (х + 2)² = 1

х + 2 = 1 или х + 2 = - 1

х = - 1 или х = - 3

2) (х + 2)² = 4

х + 2 = 2 или х + 2 = - 2

х = 0 или х = - 4

Ответ: -4; -3; -1; 0.

Answer 2

To solve the equation $(x+2)^4 - 5(x+2)^2 + 4 = 0$, you can use a substitution to simplify it and then solve for the variable $x$. Let $y = (x+2)^2$, so the equation becomes:

$y^2 - 5y + 4 = 0$

Now, you can solve this quadratic equation for $y$:

$y^2 - 5y + 4 = 0$

This quadratic equation can be factored as:

$(y - 1)(y - 4) = 0$

Now, you can set each factor equal to zero and solve for $y$:

1. $y - 1 = 0$ implies $y = 1$
2. $y - 4 = 0$ implies $y = 4$

Now, remember that $y = (x+2)^2$. So, you have two possibilities:

1. $(x+2)^2 = 1$
2. $(x+2)^2 = 4$

Now, solve for $x$ in each case:

1. For $(x+2)^2 = 1$, you have two cases: a. $x+2 = 1$, which gives $x = -1$. b. $x+2 = -1$, which gives $x = -3$.

2. For $(x+2)^2 = 4$, you have two cases: a. $x+2 = 2$, which gives $x = 0$. b. $x+2 = -2$, which gives $x = -4$.

So, the solutions to the original equation $(x+2)^4 - 5(x+2)^2 + 4 = 0$ are $x = -1, -3, 0, -4$.