Q. \(x^{2}-4x+4=0\)

Q: What are the solutions to x^2-4x+4=0?

x^2-4x+4=(x-2)^2, so x-2=0\Rightarrow x=2 (double root).

Q: How do I factor x^2-4x+4?

Recognize a perfect square: x^2-4x+4=x^2-2\cdot 2\cdot x+2^2=(x-2)^2.

Q: Use the quadratic formula to solve x^2-4x+4=0.

a=1,b=-4,c=4. Then x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-16}}{2}=2.

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

D=b^2-4ac=(-4)^2-4(1)(4)=16-16=0. So there is one real solution of multiplicity 2.

Q: How can I solve by completing the square?

x^2-4x+4=(x^2-4x+4)=(x-2)^2. Set (x-2)^2=0\Rightarrow x=2.

Q: What is the vertex of the parabola y=x^2-4x+4, and where is it relative to the x-axis?

Vertex at x=-\frac{b}{2a}=2. Value y=0, so it touches the x-axis at x=2 (tangent).

Answer

We solve the quadratic \(x^2-4x+4=0\). This factors as \(x^2-4x+4=(x-2)^2\).

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

So \(x-2=0\), giving \(x=2\).

Final result: \(x=2\) (a double root).

Detailed Explanation

We want to solve the equation:

\[
x^2 – 4x + 4 = 0
\]

Step 1: Recognize the quadratic form

The expression \(x^2 – 4x + 4\) looks like a perfect square because:

\(x^2\) matches the first term of \((x – \text{something})^2\)
\(-4x\) suggests the “something” is \(2\), since \(-2 \cdot 2x = -4x\)
\(4\) matches \(2^2 = 4\)

Step 2: Factor using the perfect-square pattern

Use the identity:

\[
(a-b)^2 = a^2 – 2ab + b^2
\]

Let \(a = x\) and \(b = 2\). Then:

\[
(x-2)^2 = x^2 – 2(x)(2) + 2^2 = x^2 – 4x + 4
\]

So the equation becomes:

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

Step 3: Solve by taking the square root

If \((x-2)^2 = 0\), then the only way a square equals \(0\) is if the inside is \(0\):

\[
x – 2 = 0
\]

Step 4: Solve for \(x\)

Add \(2\) to both sides:

\[
x = 2
\]

Final Answer

The solution is:

\[
x = 2
\]

See full solution

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Algebra FAQ

What are the solutions to \(x^2-4x+4=0\)?

\(x^2-4x+4=(x-2)^2\), so \(x-2=0\Rightarrow x=2\) (double root).

How do I factor \(x^2-4x+4\)?

Recognize a perfect square: \(x^2-4x+4=x^2-2\cdot 2\cdot x+2^2=(x-2)^2\).

Use the quadratic formula to solve \(x^2-4x+4=0\).

\(a=1,b=-4,c=4\). Then \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{4\pm\sqrt{16-16}}{2}=2\).

What is the discriminant, and what does it mean?

\(D=b^2-4ac=(-4)^2-4(1)(4)=16-16=0\). So there is one real solution of multiplicity \(2\).

How can I solve by completing the square?

\(x^2-4x+4=(x^2-4x+4)=(x-2)^2\). Set \((x-2)^2=0\Rightarrow x=2\).

What is the vertex of the parabola \(y=x^2-4x+4\), and where is it relative to the x-axis?

Vertex at \(x=-\frac{b}{2a}=2\). Value \(y=0\), so it touches the x-axis at \(x=2\) (tangent).

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