Square of sum: Domain of definition

To solve the equation $\frac{3}{(x+1)^2}+\frac{2x}{x+1}+x+1=3$, we will clear the fractions by finding a common denominator.

• Step 1: The common denominator of the fractions $\frac{3}{(x+1)^2}$ and $\frac{2x}{x+1}$ is $(x+1)^2$.
• Step 2: Multiply each term in the equation by $(x+1)^2$ to clear the fractions:
  $\left(\frac{3}{(x+1)^2} \cdot (x+1)^2\right) + \left(\frac{2x}{x+1} \cdot (x+1)^2\right) + (x+1) \cdot (x+1)^2 = 3 \cdot (x+1)^2$.
• Step 3: Simplify each term:
  - The first term becomes $3$.
  - The second term becomes $2x(x+1)$.
  - The third term becomes $(x+1)^3$.
  Then, equate to the right-hand side: $3 + 2x(x+1) + (x+1)^3 = 3(x+1)^2$.
• Step 4: Expand the expressions:
  - Expand $2x(x+1)$ to get $2x^2 + 2x$.
  - Expand $(x+1)^3$ to $x^3 + 3x^2 + 3x + 1$.
  - Expand $3(x+1)^2$ to $3(x^2 + 2x + 1)$ or $3x^2 + 6x + 3$.
• Step 5: Formulate the new equation by bringing all terms to one side:
  $x^3 + 3x^2 + 3x + 1 + 2x^2 + 2x + 3 - 3x^2 - 6x - 3 = 0$.
• Combine like terms to simplify:
  $x^3 + 2x^2 - x + 1 = 0$
• Step 6: Solve the resulting equation, which is already simplified:
  Factor the equation if possible. Here we substitute likely values or use a factoring method.
  Using the Rational Root Theorem or graphically analyzing roots might give viable real solutions.
  Let's factor even further:
  $(x - (\sqrt{3}-2))(x - (-\sqrt{3}-2)) = 0$.
• Step 7: Solve for $x$ from the factors:
  $x = \sqrt{3} - 2$ or $x = -\sqrt{3} - 2$.

Thus, the values of $x$ that satisfy this equation are $x = \sqrt{3} - 2$ and $x = -\sqrt{3} - 2$.

Therefore, the correct choice is:

$x = \sqrt{3} - 2, -\sqrt{3} - 2$

To solve the equation $\frac{x^3 + 1}{(x+1)^2} = x$, we will follow these steps:

• Step 1: Set up the equation for solving by cross-multiplying.
• Step 2: Simplify and solve the resulting polynomial equation.
• Step 3: Solve for the values of $x$.

Let's work through the solution:

Step 1: Cross-multiply to eliminate the fraction:

Expand the right-hand side:

Step 2: Set the expanded equation equal:

Cancel $x^3$ from both sides:

Re-arrange the equation to form a standard quadratic equation:

Step 3: Solve the quadratic equation using the quadratic formula:

The quadratic formula is:

Substitute the values of $a$, $b$, and $c$ into the formula:

Calculate the discriminant and simplify:

Simplify further:

This gives the solutions:

Since $x = -1$ would make the denominator zero, it is not allowed as a solution. Thus, the only valid solution is:

Therefore, the solution to the equation is $x = \frac{1}{2}$.

In order to solve the equation, start by removing the denominators.

To do this, we'll multiply the denominators:

$(2x+1)^2\cdot(2x+1)+(x+2)^2\cdot(x+2)=4.5x(2x+1)(x+2)$

Open the parentheses on the left side, making use of the distributive property:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x+1)(x+2)$

Continue to open the parentheses on the right side of the equation:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=4.5x(2x^2+5x+2)$

Simplify further:

$(4x^2+4x+1)\cdot(2x+1)+(x^2+4x+4)\cdot(x+2)=9x^3+22.5x+9x$

Go back and simplify the parentheses on the left side of the equation:

$8x^3+8x^2+2x+4x^2+4x+1+x^3+4x^2+4x+2x^2+8x+8=9x^3+22.5x+9x$

Combine like terms:

$9x^3+18x^2+18x+9=9x^3+22.5x+9x$

Notice that all terms can be divided by 9 as shown below:

$x^3+2x^2+2x+1=x^3+2.5x+x$

Move all numbers to one side:

$x^3-x^3+2x^2-2.5x^2+2x-x+9=0$

We obtain the following:

$0.5x^2-x-1=0$

In order to remove the one-half coefficient, multiply the entire equation by 2

$x^2-2x-2=0$

Apply the square root formula, as shown below-

$\frac{2±\sqrt{12}}{2}$

Apply the properties of square roots in order to simplify the square root of 12:

$\frac{2±2\sqrt{3}}{2}$Divide both the numerator and denominator by 2 as follows:

$1±\sqrt{3}$