Question

(x+3)⁴ - 3(x+3)² + 2=0

Answer 1

В приложении ...........

Answer 2

(x+3)⁴ - 3(x+3)² + 2=0
Пусть: (х+3)^2=а
Тогда: а^2-3а+2=0
По теореме Виета:
а1+а2=-(-3)=3
а1*а2=2
а1=1
а2=2

(х1+3)^2=а1=1
х1^2+6х1+9=1
х1^2+6х1+9-1=0
х1^2+6х1+8=0
По теореме Виета:
х1(1)+х1(2)=-6
х1(1)*х1(2)=8
х1(1)=-4
х1(2)=-2

(х2+3)^2=а2=2
х2^2+6х2+9=2
х2^2+6х2+9-2=0
х2^2+6х2+7=0
D=(-6)^2-4*1*7=36-28=8
x2(1)=(-6-V8)/2*1=-3-(V8/2)=-3-V2
x2(2)=(-6+V8)/2=-3+(V8/2)=-3+V2

Answer 3

To solve the equation $(x+3)^4 - 3(x+3)^2 + 2 = 0$, we can use a substitution to simplify it and then solve for the unknown variable $x$. Let's make a substitution:

Let $y = (x+3)^2$.

Now, our equation becomes:

$y^2 - 3y + 2 = 0$.

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

$y^2 - 3y + 2 = (y-1)(y-2) = 0$.

Setting each factor to zero and solving for $y$:

1. $y - 1 = 0$ gives $y = 1$.
2. $y - 2 = 0$ gives $y = 2$.

Now, recall that $y = (x+3)^2$. We can use these values of $y$ to find the corresponding values of $x$:

1. When $y = 1$: $1 = (x + 3)^2$. Taking the square root of both sides, we get: $x + 3 = \pm \sqrt{1}$. $x + 3 = \pm 1$. Solving for $x$: $x = -3 + 1 = -2$ or $x = -3 - 1 = -4$.

2. When $y = 2$: $2 = (x + 3)^2$. Taking the square root of both sides, we get: $x + 3 = \pm \sqrt{2}$. $x + 3 = \pm \sqrt{2}$. Solving for $x$: $x = -3 + \sqrt{2}$ or $x = -3 - \sqrt{2}$.

So, the solutions for the original equation are $x = -2$, $x = -4$, $x = -3 + \sqrt{2}$, and $x = -3 - \sqrt{2}$.