Q. \(x^{2}+1=0\)

Q: How do you use the square root to solve x^2+1=0? \\

From x^2=-1, take square roots: x=\sqrt{-1}=\pm i.

Q: Why are there no real solutions to x^2+1=0? \\

For real x, x^2\ge 0 so x^2+1\ge 1, never 0.

Q: Can you solve it using the quadratic formula? \\

a=1,b=0,c=1. Then x=\frac{-0\pm\sqrt{0-4}}{2}=\pm i.

Q: What is the discriminant of x^2+1=0? \\

\Delta=b^2-4ac=0-4\cdot 1\cdot 1=-4, so complex roots occur.

Q: What does x^2=-1 mean in complex numbers? \\

It means x must be imaginary since i^2=-1, giving x=\pm i.

Answer

We solve \(x^2+1=0\) by moving \(1\) to the other side:

\(x^2=-1\).

Since \( -1 = i^2 \), we take square roots:

\(x=\pm i\).

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

Detailed Explanation

We want to solve the equation

\[
x^2 + 1 = 0
\]

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

Subtract \(1\) from both sides so that the \(x^2\) term is by itself.

\[
x^2 + 1 – 1 = 0 – 1
\]
\[
x^2 = -1
\]

Step 2: Take the square root of both sides.

By the square root rule, we write both the positive and negative square roots.

\[
x = \pm \sqrt{-1}
\]

Step 3: Use the definition of the imaginary unit.

We know that

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

So the solutions become

\[
x = \pm i
\]

Final Answer:

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

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Algebra FAQ

What are the solutions to \(x^2+1=0\)? \\

Solve \(x^2=-1\), so \(x=\pm i\).

How do you use the square root to solve \(x^2+1=0\)? \\

From \(x^2=-1\), take square roots: \(x=\sqrt{-1}=\pm i\).

Why are there no real solutions to \(x^2+1=0\)? \\

For real \(x\), \(x^2\ge 0\) so \(x^2+1\ge 1\), never \(0\).

Can you solve it using the quadratic formula? \\

\(a=1,b=0,c=1\). Then \(x=\frac{-0\pm\sqrt{0-4}}{2}=\pm i\).

What is the discriminant of \(x^2+1=0\)? \\

\(\Delta=b^2-4ac=0-4\cdot 1\cdot 1=-4\), so complex roots occur.

What does \(x^2=-1\) mean in complex numbers? \\

It means \(x\) must be imaginary since \(i^2=-1\), giving \(x=\pm i\).

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