Q. \((x-7)^2\)
Answer
To simplify \((x-7)^2\), use the identity \((a-b)^2=a^2-2ab+b^2\). Here \(a=x\) and \(b=7\).
\[
(x-7)^2=x^2-2\cdot x \cdot 7+7^2=x^2-14x+49
\]
Detailed Explanation
We want to expand the expression \( (x-7)^2 \).
Step 1: Use the square of a binomial rule.
The formula for squaring a binomial is:
\[
(a-b)^2=a^2-2ab+b^2
\]
Here, \(a=x\) and \(b=7\), so:
\[
(x-7)^2=x^2-2(x)(7)+7^2
\]
Step 2: Compute each part.
First, compute the middle term:
\[
-2(x)(7)=-14x
\]
Next, compute \(7^2\):
\[
7^2=49
\]
Step 3: Combine everything.
Substitute the results back:
\[
(x-7)^2=x^2-14x+49
\]
Final Answer:
\[
(x-7)^2=x^2-14x+49
\]
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Algebra FAQ
What is the expanded form of \( (x-7)^2 \)?
\( (x-7)^2 = x^2 - 14x + 49 \).
How can I expand \( (x-7)^2 \) using the FOIL method?
Multiply: \(x\cdot x=x^2\), \(x\cdot (-7)=-7x\), \((-7)\cdot x=-7x\), \((-7)\cdot(-7)=49\). Add: \(x^2-14x+49\).
What is the vertex of \( y=(x-7)^2+0 \)?
Since \(y=(x-h)^2+k\) has vertex \((h,k)\), the vertex is \( (7,0) \).
For which \(x\) does \( (x-7)^2 = 0 \)?
A square equals \(0\) only when the inside is \(0\): \(x-7=0\), so \(x=7\).
Solve \( (x-7)^2 = 9 \).
Take square roots: \(x-7=\pm 3\). So \(x=10\) or \(x=4\).
Is \( (x-7)^2 \) always nonnegative?
Yes. For any real \(x\), \( (x-7)^2 \ge 0 \) since it is a square.
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