Q. \(x^2 – 2x + 1 = 0\)

Q: What are the solutions of x^2-2x+1=0?

\left(x-1\right)^2=0, so x=1 (double root).

Q: What does the discriminant tell us?

For ax^2+bx+c: \Delta=b^2-4ac=(-2)^2-4(1)(1)=0, so one repeated real root.

Q: How do I solve it by completing the square?

x^2-2x+1=(x^2-2x+1)=\left(x-1\right)^2=0, hence x=1.

Q: What is the vertex and how it relates to the root?

f(x)=x^2-2x+1=(x-1)^2 has vertex at x=1, value 0, so it touches the x-axis there.

Answer

We solve the quadratic equation

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

Factor the left side:

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

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

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

Detailed Explanation

We want to solve the equation

\[
x^2 – 2x + 1 = 0.
\]

Step 1: Notice the expression is a perfect square.

The left side has the form

\[
x^2 – 2x + 1.
\]

Compare this with the perfect square pattern

\[
(a-b)^2 = a^2 – 2ab + b^2.
\]

Here, we can match

\[
a = x,\quad b = 1.
\]

Then

\[
(x-1)^2 = x^2 – 2x + 1.
\]

Step 2: Rewrite the equation using the perfect square.

\[
x^2 – 2x + 1 = 0
\]

becomes

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

Step 3: Use the zero-product (or square-equals-zero) idea.

\[
(x-1)^2 = 0,
\]

then the squared quantity must be zero:

\[
x – 1 = 0.
\]

Step 4: Solve for \(x\).

\[
x – 1 = 0
\]
\[
x = 1.
\]

Final Answer:

\[
x = 1.
\]

See full solution

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Algebra FAQ

What are the solutions of \(x^2-2x+1=0\)?

\(\left(x-1\right)^2=0\), so \(x=1\) (double root).

How can I factor \(x^2-2x+1\)?

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

What does the discriminant tell us?

For \(ax^2+bx+c\): \(\Delta=b^2-4ac=(-2)^2-4(1)(1)=0\), so one repeated real root.

How do I solve it by completing the square?

\(x^2-2x+1=(x^2-2x+1)=\left(x-1\right)^2=0\), hence \(x=1\).

What is the vertex and how it relates to the root?

\(f(x)=x^2-2x+1=(x-1)^2\) has vertex at \(x=1\), value \(0\), so it touches the \(x\)-axis there.

How can I use the quadratic formula?

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{2\pm\sqrt{0}}{2\cdot1}=1\).

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