Q. \(x^2 = 1\)

Q: Why does x^2=1 imply x=\pm 1?

Taking square roots gives x=\sqrt{1} or x=-\sqrt{1}. Since \sqrt{1}=1, solutions are x=1 and x=-1.

Q: Solve x^2=1 using the square root method carefully.

From x^2=1, write x=\pm\sqrt{1}. Then x=\pm 1, so x=1 and x=-1.

Q: What number set are the solutions in (real vs complex)?

Over reals, solutions are 1 and -1. Over complex numbers, no new solutions appear because the factorization (x-1)(x+1)=0 still holds.

Q: How would the solution change if x^2=-1?

Over reals, there are no solutions. Over complex numbers, x=\pm i, since i^2=-1.

Q: Is x=0 a solution to x^2=1?

No. Substituting gives 0^2=0\neq 1. Therefore x=0 is not a solution.

Answer

We solve the equation \(x^2=1\). Taking square roots gives \(x=\pm 1\).

\[
x = 1 \text{ or } x = -1
\]

Detailed Explanation

We are asked to solve the equation

\[
x^2 = 1
\]

Step 1: Take square roots of both sides.

Since \(x^2\) is a square, the solutions come from

\[
x = \pm \sqrt{1}
\]

Step 2: Simplify the square root.

\[
\sqrt{1} = 1
\]

\[
x = \pm 1
\]

Step 3: Write both solutions explicitly.

\[
x = 1 \quad \text{or} \quad x = -1
\]

Final Answer: \(\,x = 1\) or \(\,x = -1\).

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Algebra FAQ

What are all solutions to \(x^2=1\)?

\(x^2-1=0\Rightarrow (x-1)(x+1)=0\Rightarrow x=1 \text{ or } x=-1\).

Why does \(x^2=1\) imply \(x=\pm 1\)?

Taking square roots gives \(x=\sqrt{1}\) or \(x=-\sqrt{1}\). Since \(\sqrt{1}=1\), solutions are \(x=1\) and \(x=-1\).

Solve \(x^2=1\) using the square root method carefully.

From \(x^2=1\), write \(x=\pm\sqrt{1}\). Then \(x=\pm 1\), so \(x=1\) and \(x=-1\).

What number set are the solutions in (real vs complex)?

Over reals, solutions are \(1\) and \(-1\). Over complex numbers, no new solutions appear because the factorization \((x-1)(x+1)=0\) still holds.

How would the solution change if \(x^2=-1\)?

Over reals, there are no solutions. Over complex numbers, \(x=\pm i\), since \(i^2=-1\).

Is \(x=0\) a solution to \(x^2=1\)?

No. Substituting gives \(0^2=0\neq 1\). Therefore \(x=0\) is not a solution.

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