Q. \(x^2 + 49 = 0\)

Q: How do you use factoring for x^2+49=0?

Write x^2+49=x^2+(7i)^2=(x-7i)(x+7i)=0. Then x=\pm 7i.

Q: Can you solve it using the square root method?

From x^2=-49, take square roots: x=\sqrt{-49}=\pm\sqrt{49}\,i=\pm 7i.

Q: What is the discriminant for x^2+49=0 and what does it imply?

For ax^2+bx+c=0, a=1,b=0,c=49. Discriminant D=b^2-4ac=0-196=-196<0, so two complex conjugate roots.

Q: Is there any real solution to x^2+49=0?

No. Since x^2\ge 0, x^2+49>0 for all real x. Thus no real roots.

Answer

We solve \(x^2+49=0\):

\[
x^2=-49
\]
\[
x=\pm\sqrt{-49}=\pm 7i
\]

Final result: \(x=7i\) or \(x=-7i\).

Detailed Explanation

We want to solve the equation \(x^2+49=0\).

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

The equation is

\[
x^2+49=0
\]

Subtract \(49\) from both sides to get \(x^2\) by itself:

\[
x^2+49-49=0-49
\]

This simplifies to

\[
x^2=-49
\]

Step 2: Solve for \(x\).

Now take the square root of both sides:

\[
x=\pm \sqrt{-49}
\]

Step 3: Rewrite \(\sqrt{-49}\) using \(i\).

Recall that \(i=\sqrt{-1}\), so \(\sqrt{-49}=\sqrt{49}\sqrt{-1}=7i\). Therefore:

\[
x=\pm 7i
\]

Final Answer:

\[
x=\pm 7i
\]

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Algebra FAQ

Solve \(x^2+49=0\).

\(x^2=-49\). So \(x=\pm 7i\).

What are the complex roots of \(x^2+49=0\)?

The roots are \(x=7i\) and \(x=-7i\).

How do you use factoring for \(x^2+49=0\)?

Write \(x^2+49=x^2+(7i)^2=(x-7i)(x+7i)=0\). Then \(x=\pm 7i\).

Can you solve it using the square root method?

From \(x^2=-49\), take square roots: \(x=\sqrt{-49}=\pm\sqrt{49}\,i=\pm 7i\).

What is the discriminant for \(x^2+49=0\) and what does it imply?

For \(ax^2+bx+c=0\), \(a=1,b=0,c=49\). Discriminant \(D=b^2-4ac=0-196=-196<0\), so two complex conjugate roots.

Is there any real solution to \(x^2+49=0\)?

No. Since \(x^2\ge 0\), \(x^2+49>0\) for all real \(x\). Thus no real roots.

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