Q. \(x^2 + 2x + 1 = 0\)

Q: Solve x^2+2x+1=0 by factoring.

(x+1)^2=0\Rightarrow x=-1 (double root).

Q: Solve using the quadratic formula.

For a=1,b=2,c=1: x=\frac{-2\pm\sqrt{4-4}}{2}=-1.

Q: What is the discriminant and what does it mean?

D=b^2-4ac=4-4=0, so there is one real repeated root x=-1.

Q: How do you complete the square?

x^2+2x+1=(x+1)^2. Set to zero: (x+1)^2=0\Rightarrow x=-1.

Q: What are the multiplicity and number of solutions?

Multiplicity is 2 for root x=-1. Number of distinct real solutions: 1.

Answer

We solve \(x^2+2x+1=0\). Note that \(x^2+2x+1=(x+1)^2\).

\[
(x+1)^2=0
\]
So \(x+1=0\), hence \(x=-1\).

Final result: \(x=-1\) (double root).

Detailed Explanation

We want to solve the equation

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

Step 1: Recognize the expression as a perfect square.

Notice that

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

matches the structure

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

So we can rewrite the equation as

\[
(x+1)^2=0.
\]

Step 2: Use the zero product idea.

If a square equals zero, then the inside must be zero. That means

\[
x+1=0.
\]

Step 3: Solve for \(x\).

Subtract \(1\) from both sides:

\[
x=-1.
\]

Final Answer:

\[
x=-1.
\]

See full solution

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Algebra FAQ

How do you factor \(x^2+2x+1\)?

\(x^2+2x+1=(x+1)^2\).

Solve \(x^2+2x+1=0\) by factoring.

\((x+1)^2=0\Rightarrow x=-1\) (double root).

Solve using the quadratic formula.

For \(a=1,b=2,c=1\): \(x=\frac{-2\pm\sqrt{4-4}}{2}=-1\).

What is the discriminant and what does it mean?

\(D=b^2-4ac=4-4=0\), so there is one real repeated root \(x=-1\).

How do you complete the square?

\(x^2+2x+1=(x+1)^2\). Set to zero: \((x+1)^2=0\Rightarrow x=-1\).

What are the multiplicity and number of solutions?

Multiplicity is \(2\) for root \(x=-1\). Number of distinct real solutions: \(1\).

Use tools to solve the equation.
Check roots: (x+1)^2=0.

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