Q. \(x^2-6x+9\)
Answer
Recognize the quadratic as a perfect square trinomial.
Rewrite:
\[
x^2 – 6x + 9 = (x-3)^2
\]
Final result:
\[
(x-3)^2
\]
Detailed Explanation
We want to simplify the expression \(x^2 – 6x + 9\) by rewriting it in a more useful form.
Step 1: Recognize the expression as a perfect square pattern.
A common perfect square trinomial looks like:
\[
a^2 – 2ab + b^2 = (a-b)^2
\]
In this form, the middle term is \(-2ab\).
Step 2: Match coefficients to our expression.
Our expression is:
\[
x^2 – 6x + 9
\]
Compare term by term with \(a^2 – 2ab + b^2\):
- \(a^2\) corresponds to \(x^2\), so \(a = x\).
- \(b^2\) corresponds to \(9\), so \(b = 3\) (since \(3^2 = 9\)).
- Check the middle term: \(-2ab = -2(x)(3) = -6x\), which matches perfectly.
Step 3: Rewrite as a perfect square.
Since it matches the pattern \(a^2 – 2ab + b^2 = (a-b)^2\), we substitute \(a = x\) and \(b = 3\):
\[
x^2 – 6x + 9 = (x-3)^2
\]
Final Answer:
\[
x^2 – 6x + 9 = (x-3)^2
\]
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Algebra FAQ
What is the simplified form of \(x^2-6x+9\)?
It is already simplified: \(x^2-6x+9\).
Can you factor \(x^2-6x+9\)?
It factors as a perfect square: \(x^2-6x+9=(x-3)^2\).
What is the vertex of the parabola \(y=x^2-6x+9\)?
Using \(y=(x-3)^2\), the vertex is at \((3,0)\).
Solve \(x^2-6x+9=0\).
\( (x-3)^2=0 \Rightarrow x=3\) (a repeated root).
What are the roots and their multiplicities?
The only root is \(x=3\) with multiplicity \(2\).
What is the y-intercept of \(y=x^2-6x+9\)?
Substitute \(x=0\): \(y=9\). So the intercept is \((0,9)\).
What is the minimum value of \(y\)?
Since \(y=(x-3)^2\ge 0\), the minimum value is \(0\) at \(x=3\).
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