Question

(x+1)^2+√x+1=0 СРОООЧНО!!

Answer 1

Замена

$\sqrt{x+1}=t$. Получаем

t⁴+t=0
t(t+1)(t²-t+1)=0
t₁=0; t₂=-1
t₂-t+1=0 - нет решения. Так как D=(-1)²-4=-3<0.
При t₁=0
х+1=0
х=-1 - решение подходит.

При t₂=-1
$\sqrt{x+1}=-1$

Тоже нет решения. Так как подкоренное выражение не должно быть отрицательным.

Ответ: х=-1.

Answer 2

To solve the equation $(x+1)^2 + \sqrt{x+1} = 0$, follow these steps:

Step 1: Identify the radical term The radical term in the equation is $\sqrt{x+1}$.

Step 2: Isolate the radical term on one side of the equation Move the $(x+1)^2$ term to the other side of the equation: $\sqrt{x+1} = -(x+1)^2$

Step 3: Square both sides of the equation Squaring both sides of the equation to eliminate the square root: $(\sqrt{x+1})^2 = ((x+1)^2)^2$ $x+1 = (x+1)^4$

Step 4: Simplify the equation Expand the right-hand side: $x+1 = (x^2 + 2x + 1)^2$ $x+1 = (x^4 + 4x^3 + 6x^2 + 4x + 1)$

Step 5: Move all terms to one side of the equation Subtract $x+1$ from both sides: $0 = x^4 + 4x^3 + 6x^2 + 4x + 1 - (x+1)$ $0 = x^4 + 4x^3 + 6x^2 + 3x$

Step 6: Factor the equation (if possible) Unfortunately, this equation is not easily factorizable. We'll try to find the solutions through numerical methods.

Step 7: Find the solutions Using numerical methods (e.g., numerical approximation or graphical methods), you can find the solutions for this equation. It's not straightforward to express the exact solutions in a simple form due to the quartic nature of the equation.

If you need an approximate solution, you can use numerical techniques like the Newton-Raphson method or numerical software to find the roots.

Keep in mind that this equation might have complex solutions since it involves a square root. It's essential to verify the solutions found in the original equation to ensure they satisfy the equation.