Q. \( x^2 = -4 \)

Q: Solve x^2=-4 over the complex numbers.\n

Take square roots: x=\pm\sqrt{-4}=\pm 2i.

Q: Does x^2=-4 have real solutions?\n

No. A real square satisfies x^2\ge 0, so it cannot equal -4.

Q: What are the principal square roots of -4?\n

\sqrt{-4}=2i (principal branch), so the square roots are 2i and -2i.

Q: Solve x^2+4=0 using the quadratic formula.\n

x=\frac{-0\pm\sqrt{0-16}}{2}=\pm\sqrt{-4}=\pm 2i.

Q: How do you rewrite -4 in polar form to find square roots?\n

-4=4e^{i\pi} (or 4e^{i(\pi+2k\pi)}). Then \sqrt{-4}=2e^{i\pi/2}=\pm 2i.

Answer

We solve \(x^2=-4\). Since \(x\) is real only if \(x^2\ge 0\), there are no real solutions. Over the complex numbers, write \(-4=(2i)^2\) so

\[
x=\pm 2i.
\]

Detailed Explanation

We are asked to solve the equation \(x^2=-4\).

Step 1: Understand what kind of numbers can satisfy the equation.

The left side is a square, so we should think about whether real numbers can work. For real \(x\), the value \(x^2\) is always greater than or equal to \(0\). That means a real number cannot produce \(-4\).

So, we must use complex numbers.

Step 2: Rewrite \(-4\) in terms of \(-1\).

We can write \(-4\) as \(-4 = 4(-1)\). Using the imaginary unit \(i\), where \(i^2=-1\), we get:

\[
x^2 = -4 = 4(-1) = 4i^2
\]

Step 3: Take the square root of both sides.

From \(x^2=-4\), we take square roots:

\[
x = \pm \sqrt{-4}
\]

Step 4: Compute \(\sqrt{-4}\).

Break \(-4\) into \(-1\cdot 4\):

\[
\sqrt{-4}=\sqrt{(-1)\cdot 4}=\sqrt{-1}\sqrt{4}
\]

Now substitute \(\sqrt{-1}=i\) and \(\sqrt{4}=2\):

\[
\sqrt{-4}=2i
\]

Step 5: Include both the positive and negative square roots.

\[
x = \pm 2i
\]

Final Answer: The solutions to \(x^2=-4\) are \(x=2i\) and \(x=-2i\).

See full solution

Stuck on x² = −4? Try our AI Homework Help!

Homework AI

Algebra FAQ

Solve \(x^2=-4\) over the complex numbers.\n

Take square roots: \(x=\pm\sqrt{-4}=\pm 2i\).

Does \(x^2=-4\) have real solutions?\n

No. A real square satisfies \(x^2\ge 0\), so it cannot equal \(-4\).

What are the principal square roots of \(-4\)?\n

\(\sqrt{-4}=2i\) (principal branch), so the square roots are \(2i\) and \(-2i\).

Solve \(x^2+4=0\) using the quadratic formula.\n

\(x=\frac{-0\pm\sqrt{0-16}}{2}=\pm\sqrt{-4}=\pm 2i\).

How do you rewrite \(-4\) in polar form to find square roots?\n

\(-4=4e^{i\pi}\) (or \(4e^{i(\pi+2k\pi)}\)). Then \(\sqrt{-4}=2e^{i\pi/2}=\pm 2i\).

Why are complex numbers needed here?\n

Because no real number squared equals a negative value. Introducing \(i\) allows \(\,i^2=-1\), making \(x^2=-4\) possible.

Solve x^2=-4 step by step.
Use Math AI to check results.

298,376+ active customers
Math, Geometry, Trigonometry, etc.

Math Solver Geometry Solver Trigonometry Solver