Q. \(x^2 + 81 = 0\)

Q: Solve x^2+81=0.

x^2=-81. So x=\pm\sqrt{-81}=\pm 9i.

Q: How do you rewrite x^2+81=0 to isolate x?

Move 81: x^2=-81. Then take square roots: x=\pm\sqrt{-81}=\pm 9i.

Q: Does x^2+81=0 have real solutions?

No. For real x, x^2\ge 0, so x^2+81>0 always. Hence no real solutions.

Q: What is \sqrt{-81}?

\sqrt{-81}=\sqrt{81}\sqrt{-1}=9i. So solutions use \pm 9i.

Q: Verify the solutions x=\pm 9i in the equation.

Substitute: (9i)^2+81=81(-1)+81=0. Similarly, (-9i)^2+81=0.

Answer

We solve \(x^2+81=0\) by isolating \(x^2\):

\[
x^2+81=0
\]
\[
x^2=-81
\]

Since \(-81=81(-1)\), we have \(x^2=(9i)^2\), so \(x=\pm 9i\).

Final result: \(x=9i\) or \(x=-9i\).

Detailed Explanation

We want to solve the equation

\[
x^2 + 81 = 0
\]

Step 1: Move the constant term to the other side.

Start with

\[
x^2 + 81 = 0
\]

Subtract \(81\) from both sides so that the \(x^2\) term is alone.

\[
x^2 + 81 – 81 = 0 – 81
\]

Simplify.

\[
x^2 = -81
\]

Step 2: Take the square root of both sides.

Because \(x^2 = -81\), we write

\[
x = \pm \sqrt{-81}
\]

Step 3: Simplify the square root of a negative number.

Use the fact that \(\sqrt{-1} = i\), where \(i\) is the imaginary unit.

\[
\sqrt{-81} = \sqrt{81}\sqrt{-1} = 9i
\]

So the solutions are

\[
x = \pm 9i
\]

Final Answer:

\[
x = 9i \quad \text{or} \quad x = -9i
\]

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Algebra FAQ

Solve \(x^2+81=0\).

\(x^2=-81\). So \(x=\pm\sqrt{-81}=\pm 9i\).

What are the complex roots of \(x^2+81=0\)?

The roots are \(x=9i\) and \(x=-9i\).

How do you rewrite \(x^2+81=0\) to isolate \(x\)?

Move \(81\): \(x^2=-81\). Then take square roots: \(x=\pm\sqrt{-81}=\pm 9i\).

Does \(x^2+81=0\) have real solutions?

No. For real \(x\), \(x^2\ge 0\), so \(x^2+81>0\) always. Hence no real solutions.

What is \(\sqrt{-81}\)?

\(\sqrt{-81}=\sqrt{81}\sqrt{-1}=9i\). So solutions use \(\pm 9i\).

Verify the solutions \(x=\pm 9i\) in the equation.

Substitute: \((9i)^2+81=81(-1)+81=0\). Similarly, \((-9i)^2+81=0\).

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