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

Q: What are the solutions using the quadratic formula?

x=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}.

Q: What is the discriminant, and what does it mean?

D=b^2-4ac=1-4=-3. Negative D means two complex conjugate roots.

Q: Can I solve it by completing the square?

x^2+x+1=\left(x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}=0, so x+\tfrac{1}{2}=\pm i\tfrac{\sqrt{3}}{2}.

Q: Are the roots complex conjugates, and what are their real and imaginary parts?

Yes. Roots are \frac{-1}{2}\pm i\frac{\sqrt{3}}{2}, so real part is -\tfrac{1}{2} and imaginary parts are \pm \tfrac{\sqrt{3}}{2}.

Q: What is x^2 or 1/x for solutions of the equation?

From x^2+x+1=0, x^2=-(x+1) and (if x\neq 0) \frac{1}{x}=-x-1.

Answer

We solve the quadratic \(x^2+x+1=0\) using the quadratic formula:

\[
x=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm\sqrt{-3}}{2}
\]

\[
x=\frac{-1\pm i\sqrt{3}}{2}
\]

Final result: \(x=\frac{-1+i\sqrt{3}}{2}\) or \(x=\frac{-1-i\sqrt{3}}{2}\).

Detailed Explanation

We are asked to solve the equation

\[
x^2+x+1=0.
\]

This is a quadratic equation. The standard method is the quadratic formula.

Step 1: Identify the coefficients.

Compare

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

to the form

\[
ax^2+bx+c=0.
\]

So we have:

\[
a=1,\quad b=1,\quad c=1.
\]

Step 2: Write the quadratic formula.

\[
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\]

Step 3: Substitute \(a\), \(b\), and \(c\).

\[
x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)}.
\]

Step 4: Simplify inside the square root.

\[
1^2=1,
\]

and

\[
4(1)(1)=4.
\]

So the discriminant becomes

\[
1-4=-3.
\]

Thus

\[
x=\frac{-1\pm\sqrt{-3}}{2}.
\]

Step 5: Simplify \(\sqrt{-3}\).

Use the fact that

\[
\sqrt{-3}=i\sqrt{3},
\]

where \(i\) is the imaginary unit (\(i^2=-1\)).

So the solutions are

\[
x=\frac{-1\pm i\sqrt{3}}{2}.
\]

Final Answer:

\[
\boxed{x=\frac{-1+i\sqrt{3}}{2}\quad \text{or}\quad x=\frac{-1-i\sqrt{3}}{2}.}
\]

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Algebra FAQ

How do I factor the quadratic \(x^2+x+1\)?

It does not factor over the reals/integers since the discriminant is \(1-4=-3\lt 0\).

What are the solutions using the quadratic formula?

\(x=\frac{-1\pm\sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}\).

What is the discriminant, and what does it mean?

\(D=b^2-4ac=1-4=-3\). Negative \(D\) means two complex conjugate roots.

Can I solve it by completing the square?

\(x^2+x+1=\left(x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}=0\), so \(x+\tfrac{1}{2}=\pm i\tfrac{\sqrt{3}}{2}\).

Are the roots complex conjugates, and what are their real and imaginary parts?

Yes. Roots are \(\frac{-1}{2}\pm i\frac{\sqrt{3}}{2}\), so real part is \(-\tfrac{1}{2}\) and imaginary parts are \(\pm \tfrac{\sqrt{3}}{2}\).

What is \(x^2\) or \(1/x\) for solutions of the equation?

From \(x^2+x+1=0\), \(x^2=-(x+1)\) and (if \(x\neq 0\)) \(\frac{1}{x}=-x-1\).

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