Q. \(x^2 – 64 = 0\)
Answer
We solve
\[
x^2 – 64 = 0
\]
Factor (difference of squares):
\[
x^2 – 8^2 = 0 \Rightarrow (x-8)(x+8)=0
\]
So \(x=8\) or \(x=-8\).
Final answer: \(x=8\) or \(x=-8\).
Detailed Explanation
We want to solve the equation:
\[
x^2 – 64 = 0
\]
Step 1: Move the constant term.
Start with:
\[
x^2 – 64 = 0
\]
Add \(64\) to both sides so that the \(x^2\) term is alone:
\[
x^2 – 64 + 64 = 0 + 64
\]
This simplifies to:
\[
x^2 = 64
\]
Step 2: Take the square root of both sides.
Since \(x^2 = 64\), we take square roots of both sides:
\[
x = \pm\sqrt{64}
\]
Step 3: Simplify the square root.
Note that:
\[
\sqrt{64} = 8
\]
So:
\[
x = \pm 8
\]
Answer:
\[
x = 8 \quad \text{or} \quad x = -8
\]
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Algebra FAQ
Solve \(x^2-64=0\).
\(x^2=64\), so \(x=\pm 8\).
How do you factor \(x^2-64\)?
\(x^2-64=(x-8)(x+8)\).
Solve using the square root method.
From \(x^2=64\), take roots: \(x=\sqrt{64}=8\) and \(x=-\sqrt{64}=-8\).
What is the discriminant for \(x^2-64=0\)?
For \(ax^2+bx+c=0\) with \(a=1,b=0,c=-64\), \(D=b^2-4ac=0-4(1)(-64)=256\).
Does \(x^2-64=0\) have real solutions?
Yes. Because \(D=256>0\), there are two real solutions: \(x=\pm 8\).
What are the roots of \(x^2=64\) directly?
The roots are \(x=8\) and \(x=-8\).
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