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

Answer

\(x^2-4x+4\) is a perfect square:

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

So the factorization is \((x-2)(x-2)\), and the vertex/double root is at \(x=2\).

Detailed Explanation

We are given the expression \(x^2 – 4x + 4\). The goal is typically to rewrite it in a simpler factored or completed-square form.

Step 1: Recognize a perfect square pattern

Notice that the first term is \(x^2\), and the last term is \(4\), which is \(2^2\). Also, the middle term is \(-4x\), which matches \(-2 \cdot x \cdot 2\). This suggests the expression is a perfect square of the form:

\[
x^2 – 4x + 4 = (x – 2)^2
\]

Step 2: Verify by expanding

Expand \((x – 2)^2\) to confirm it matches the original expression.

\[
(x – 2)^2 = (x – 2)(x – 2)
\]

Distribute each term:

\[
(x – 2)(x – 2) = x(x – 2) – 2(x – 2)
\]

Now distribute again:

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

Add the results:

\[
x^2 – 2x – 2x + 4 = x^2 – 4x + 4
\]

Conclusion

Since the expansion matches the original expression, we conclude:

\[
x^2 – 4x + 4 = (x – 2)^2
\]

See full solution

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Algebra FAQ

Factor \(x^2-4x+4\)?

\(\bigl(x-2\bigr)^2\).

Complete the square for \(x^2-4x+4\)?

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

Find the roots of \(x^2-4x+4=0\)?

\(\bigl(x-2\bigr)^2=0\Rightarrow x=2\) (double root).

What is the vertex and minimum value of \(y=x^2-4x+4\)?

Vertex at \((2,0)\). Minimum value is \(0\).

Expand \((x-2)^2\) to match \(x^2-4x+4\)?

\((x-2)^2=x^2-4x+4\).

Determine whether \(x^2-4x+4\) is always nonnegative?

Yes, since \(x^2-4x+4=(x-2)^2\ge 0\) for all real \(x\).
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