Q. \(x^2+36=0\)

Q: How do I rewrite the equation using subtraction?

From x^2+36=0, subtract 36: x^2=-36.

Q: How can I take the square root of a negative number?

Use \sqrt{-1}=i . Thus x=\pm \sqrt{-36}=\pm 6i.

Q: What does the quadratic formula give for x^2+36=0?

With a=1,b=0,c=36: x=\frac{-0\pm\sqrt{0-4\cdot1\cdot36}}{2}=\pm 6i.

Q: Are there real solutions to x^2+36=0?

No. Since x^2=-36 is negative, there are no real x. Solutions are complex.

Q: What is the discriminant and what does it imply?

For x^2+0x+36=0, D=0^2-4(1)(36)=-144<0. So two non-real complex solutions.

Answer

We solve the quadratic equation \(x^2+36=0\).

\(x^2=-36\), so \(x=\pm\sqrt{-36}=\pm 6i\).

\[\boxed{x=6i \text{ or } x=-6i}\]

Detailed Explanation

We want to solve the equation \(x^2+36=0\).

Step 1: Isolate the \(x^2\) term.

Start with:

\[
x^2+36=0
\]

Subtract \(36\) from both sides to get only the \(x^2\) term on the left:

\[
x^2+36-36=0-36
\]

Simplify:

\[
x^2=-36
\]

Step 2: Take the square root of both sides.

Now solve \(x^2=-36\) by taking square roots:

\[
x=\pm\sqrt{-36}
\]

Step 3: Rewrite \(\sqrt{-36}\) using \(i\).

Since \(-36=36(-1)\), we have:

\[
\sqrt{-36}=\sqrt{36}\sqrt{-1}=6i
\]

So:

\[
x=\pm 6i
\]

Final Answer:

\[
\boxed{x=6i \quad \text{or} \quad x=-6i}
\]

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Algebra FAQ

What are the solutions to \(x^2+36=0\)?

Solve \(x^2=-36\). So \(x=\pm 6i\).

How do I rewrite the equation using subtraction?

From \(x^2+36=0\), subtract \(36\): \(x^2=-36\).

How can I take the square root of a negative number?

Use \( \sqrt{-1}=i \). Thus \(x=\pm \sqrt{-36}=\pm 6i\).

What does the quadratic formula give for \(x^2+36=0\)?

With \(a=1,b=0,c=36\): \(x=\frac{-0\pm\sqrt{0-4\cdot1\cdot36}}{2}=\pm 6i\).

Are there real solutions to \(x^2+36=0\)?

No. Since \(x^2=-36\) is negative, there are no real \(x\). Solutions are complex.

What is the discriminant and what does it imply?

For \(x^2+0x+36=0\), \(D=0^2-4(1)(36)=-144<0\). So two non-real complex solutions.

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