Q. \(x^2 = -16\)

Q: Solve x^2=-16 over the real numbers?

No real solution. Since x^2 \ge 0 for real x, it cannot equal -16.

Q: Solve x^2=-16 over the complex numbers?

x^2=-16\Rightarrow x=\pm 4i, because i^2=-1.

Q: How do you rewrite x^2=-16 as x= \ ?

Take square roots: x=\pm\sqrt{-16}=\pm\sqrt{16}\sqrt{-1}=\pm 4i.

Q: What is \sqrt{-16} in complex numbers?

\sqrt{-16} = 4i for the principal square root. The other square root is -4i.

Q: How do you factor x^2+16=0?

x^2+16=(x-4i)(x+4i)=0, so x=4i or x=-4i.

Q: Why are there two solutions to x^2=-16?

Squaring removes sign information. If x=4i, then (-4i)^2=(4i)^2=-16, giving two solutions.

Answer

We solve \(x^2=-16\). Since squares of real numbers are always \(\ge 0\), the solutions must be complex.

Rewrite \(-16\) as \(16(-1)\). Then

\[
x^2=16(-1)=16\left(i^2\right)
\]
\[
x=\pm 4i
\]

Final result: \(x=\pm 4i\).

Detailed Explanation

We are asked to solve the equation \(x^2=-16\).

Step 1: Start with the equation

We have:

\[
x^2=-16
\]

Step 2: Take the square root of both sides

To solve \(x^2=-16\), take the square root of both sides:

\[
x=\pm\sqrt{-16}
\]

Step 3: Rewrite \(-16\) using \(i\)

Recall that the imaginary unit \(i\) satisfies \(i^2=-1\). We can rewrite \(-16\) as:

\[
-16=16(-1)
\]

So:

\[
\sqrt{-16}=\sqrt{16(-1)}=\sqrt{16}\sqrt{-1}=4i
\]

Step 4: Apply the \(\pm\) sign

Substitute \(\sqrt{-16}=4i\) into \(x=\pm\sqrt{-16}\):

\[
x=\pm 4i
\]

Final Answer

The solutions are:

\[
x=4i \quad \text{or} \quad x=-4i
\]

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Algebra FAQ

Solve \(x^2=-16\) over the real numbers?

No real solution. Since \(x^2 \ge 0\) for real \(x\), it cannot equal \(-16\).

Solve \(x^2=-16\) over the complex numbers?

\(x^2=-16\Rightarrow x=\pm 4i\), because \(i^2=-1\).

How do you rewrite \(x^2=-16\) as \(x= \ ?\)

Take square roots: \(x=\pm\sqrt{-16}=\pm\sqrt{16}\sqrt{-1}=\pm 4i\).

What is \(\sqrt{-16}\) in complex numbers?

\(\sqrt{-16} = 4i\) for the principal square root. The other square root is \(-4i\).

How do you factor \(x^2+16=0\)?

\(x^2+16=(x-4i)(x+4i)=0\), so \(x=4i\) or \(x=-4i\).

Why are there two solutions to \(x^2=-16\)?

Squaring removes sign information. If \(x=4i\), then \((-4i)^2=(4i)^2=-16\), giving two solutions.

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