Q. \(x^2 = 9\)
Answer
We solve the equation \(x^2=9\) by taking square roots of both sides.
\[
x=\pm\sqrt{9}=\pm 3
\]
Final result: \(x=3\) or \(x=-3\).
Detailed Explanation
We want to solve the equation \(x^2=9\).
Step 1: Take the square root of both sides.
Since \(x^2\) is a square, we will use square roots with the plus-or-minus rule:
\[
x=\pm \sqrt{9}
\]
Step 2: Evaluate the square root.
\(\sqrt{9}=3\), so:
\[
x=\pm 3
\]
Step 3: Write both solutions.
This gives two values of \(x\):
\[
x=3 \quad \text{or} \quad x=-3
\]
Final Answer: \(\boxed{x=3}\) and \(\boxed{x=-3}\).
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Algebra FAQ
Solve \(x^2=9\).
\(x=\pm 3\).
How do you solve \(x^2=9\) using square roots?
Take square roots: \(\sqrt{x^2}= \sqrt{9}\), so \(|x|=3\), hence \(x=3\) or \(x=-3\).
Why are both \(3\) and \(-3\) solutions?
Because squaring removes the sign: \((3)^2=9\) and \((-3)^2=9\). So both values satisfy \(x^2=9\).
What is the domain of \(x^2=9\)?
\(x\) can be any real number (or complex number). There’s no restriction; solutions exist for real \(x\): \(x=\pm 3\).
Are there complex solutions to \(x^2=9\)?
Yes, but the same two complex solutions occur: \(x=3\) and \(x=-3\). No additional distinct complex roots exist.
What if the equation were \(x^2=4\)?
Similarly, \(|x|=2\), so \(x=\pm 2\).
How can you check your solutions quickly?
Substitute: \(3^2=9\) and \((-3)^2=9\). Both satisfy the original equation.
Solve x²=9 with clear steps.
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