Q. \(x^{2}=-1\)

Q: Why is x^2=-1 impossible with real numbers?

For real x, x^2 \ge 0. Since -1<0, there is no real solution.

Q: What are the complex solutions to x^2=-1?

Solve by square roots: x=\pm \sqrt{-1}=\pm i.

Q: How do I solve x^2+1=0 step-by-step?

Rewrite as x^2=-1. Take square roots: x=\pm i.

Q: Is it correct to say \sqrt{-1}=i?

Using the principal square root, \sqrt{-1}=i. Generally, square roots give \pm i.

Q: Do roots of x^2=-1 have multiplicity?

Yes. The equation factors as (x-i)(x+i)=0, giving two distinct solutions i and -i, each with multiplicity 1.

Q: What’s the geometric meaning over the complex plane?

The solutions i and -i are points on the imaginary axis at distances 1 from the origin, i.e., angles \frac{\pi}{2} and \frac{3\pi}{2}.

Answer

We are solving the equation \(x^2=-1\). Over the real numbers, there is no solution because a real number squared cannot be negative. Over the complex numbers, taking square roots gives

\[
x=\pm i
\]

Final result: \(x=\pm i\).

Detailed Explanation

We are asked to solve the equation

\[x^2=-1.\]

Step 1: Recognize the type of equation.

This is a quadratic equation in the form \(x^2 = -1\). We can solve by taking square roots on both sides.

Step 2: Take square roots carefully.

Recall that if \(x^2=a\), then the solutions are \(x=\sqrt{a}\) and \(x=-\sqrt{a}\). But here \(a=-1\), which is negative, so the solutions are complex numbers.

We use the fact that the imaginary unit \(i\) satisfies

\[i^2=-1.\]

Step 3: Compare with the given equation.

Since \(i^2=-1\), one solution is \(x=i\). Also, since \((-i)^2=(-1)^2 i^2 = -1\), the other solution is \(x=-i\).

Step 4: Write the full solution set.

The solutions to \(x^2=-1\) are

\[x=i \quad \text{or} \quad x=-i.\]

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Algebra FAQ

Why is \(x^2=-1\) impossible with real numbers?

For real \(x\), \(x^2 \ge 0\). Since \(-1<0\), there is no real solution.

What are the complex solutions to \(x^2=-1\)?

Solve by square roots: \(x=\pm \sqrt{-1}=\pm i\).

How do I solve \(x^2+1=0\) step-by-step?

Rewrite as \(x^2=-1\). Take square roots: \(x=\pm i\).

Is it correct to say \(\sqrt{-1}=i\)?

Using the principal square root, \(\sqrt{-1}=i\). Generally, square roots give \(\pm i\).

Do roots of \(x^2=-1\) have multiplicity?

Yes. The equation factors as \((x-i)(x+i)=0\), giving two distinct solutions \(i\) and \(-i\), each with multiplicity \(1\).

What’s the geometric meaning over the complex plane?

The solutions \(i\) and \(-i\) are points on the imaginary axis at distances \(1\) from the origin, i.e., angles \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\).

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