Q. \(x^2 + 16 = 0\)
Answer
We solve \(x^2+16=0\).
\(x^2=-16\), so \(x=\pm 4i\).
Final result: \(x=4i\) or \(x=-4i\).
Detailed Explanation
We want to solve the equation
\[
x^2+16=0
\]
Step 1: Isolate the \(x^2\) term.
Subtract \(16\) from both sides so that the \(x^2\) term is alone on one side.
\[
x^2+16-16=0-16
\]
This simplifies to
\[
x^2=-16
\]
Step 2: Take the square root of both sides.
From \(x^2=-16\), we know \(x\) must be imaginary because the right-hand side is negative.
\[
x=\pm\sqrt{-16}
\]
Step 3: Rewrite \(\sqrt{-16}\) using \(i\).
We use the fact that \(i=\sqrt{-1}\). Then
\[
\sqrt{-16}=\sqrt{16\cdot(-1)}=\sqrt{16}\sqrt{-1}=4i
\]
So the solutions are
\[
x=\pm 4i
\]
Final Answer.
\[
\boxed{x=4i \text{ or } x=-4i}
\]
See full solution
Algebra FAQ
What are the solutions to \(x^2+16=0\)?
Solve \(x^2=-16\), so \(x=\pm 4i\).
How do you isolate \(x^2\) in \(x^2+16=0\)?
Subtract \(16\): \(x^2=-16\).
Why are the solutions complex?
Because \(x^2=-16\) is negative, and squares of real numbers cannot be negative, so \(x\) must be imaginary.
What is the square root of \(-16\)?
\(\sqrt{-16}=4i\) (principal square root). Thus the other root is \(-4i\).
How do you rewrite the equation using factoring?
It factors as \(x^2+16=(x-4i)(x+4i)=0\), giving \(x=4i\) or \(x=-4i\).
What if I use the quadratic formula on \(x^2+16=0\)?
With \(a=1,b=0,c=16\): \(x=\frac{-0\pm\sqrt{0-64}}{2}=\pm\frac{\sqrt{-64}}{2}=\pm 4i\).
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