The quadratic formula is given by:

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

Here, the equation is $x^2 + 2x + 1 = 0$, and the coefficients are:

• $a = 1$,
• $b = 2$,
• $c = 1$.

Step 1: Plug the values into the quadratic formula

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(1)}}{2(1)}$

Step 2: Simplify the discriminant

$x = \frac{-2 \pm \sqrt{4 - 4}}{2}$ $x = \frac{-2 \pm \sqrt{0}}{2}$

Step 3: Simplify further

Since $\sqrt{0} = 0$, the formula becomes: $x = \frac{-2 \pm 0}{2}$ $x = \frac{-2}{2}$

Step 4: Final answer

$x = -1$

Thus, the solution is: $x = -1$

This is a repeated root, so the equation has only one unique solution.

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Would you like further clarification or an explanation of any step?
Here are 5 related questions to expand your understanding:

1. How does the discriminant ($b^2 - 4ac$) determine the nature of the roots of a quadratic equation?
2. How would the solution differ if the discriminant were positive or negative?
3. Can you factorize $x^2 + 2x + 1$ to verify the solution?
4. What is the geometric interpretation of a repeated root in a quadratic equation?
5. How would the solution process change for a quadratic equation without real roots?

Tip: Always check the discriminant ($b^2 - 4ac$) first to understand the type of solutions you should expect.