To solve the equation \( x + 2 = \frac{1}{1 + x} \), we'll follow these steps:

### Step 1: Eliminate the fraction
Multiply both sides by \( 1 + x \) to get rid of the fraction:

\[
(x + 2)(1 + x) = 1
\]

### Step 2: Expand the left side
Expand the left side:

\[
x(1 + x) + 2(1 + x) = x + x^2 + 2 + 2x = x^2 + 3x + 2
\]

So the equation becomes:

\[
x^2 + 3x + 2 = 1
\]

### Step 3: Move all terms to one side
Subtract 1 from both sides to set the equation to 0:

\[
x^2 + 3x + 2 - 1 = 0
\]

Simplifying gives:

\[
x^2 + 3x + 1 = 0
\]

### Step 4: Solve the quadratic equation
To solve \( x^2 + 3x + 1 = 0 \), use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \( a = 1 \), \( b = 3 \), and \( c = 1 \):

\[
x = \frac{-3 \pm \sqrt{3^2 - 4(1)(1)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 4}}{2} = \frac{-3 \pm \sqrt{5}}{2}
\]

### Step 5: Write the solutions
The solutions are:

\[
x = \frac{-3 + \sqrt{5}}{2} \quad \text{or} \quad x = \frac{-3 - \sqrt{5}}{2}
\]

These are the two possible solutions for \( x \).

Would you like further details or have any questions?

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**Relative Questions:**

1. What is the quadratic formula, and how is it derived?
2. How do you determine if a quadratic equation has real solutions?
3. Can you factorize \( x^2 + 3x + 1 \) instead of using the quadratic formula?
4. What are some applications of quadratic equations in real life?
5. How would the solution change if the original equation was \( x + 2 = \frac{2}{1 + x} \)?

**Tip:** Always check your solutions by plugging them back into the original equation to verify correctness.

Discover how to solve the algebraic equation x + 2 = 1/(1 + x) by eliminating fractions and applying the quadratic formula. This detailed step-by-step guide covers essential algebraic concepts, including the quadratic equation and its solutions, ideal for students in Grades 8-10.

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