Q. \(x^2 + 4x + 1 = 0\)

Q: How do I solve x^2+4x+1=0 using factoring?

\Delta=16-4=12, so it does not factor over integers. Use the quadratic formula: x=\frac{-4\pm\sqrt{12}}{2}=-2\pm\sqrt{3}.

Q: What is the discriminant and what does it tell me?

\Delta=b^2-4ac=4^2-4(1)(1)=12>0. So there are two distinct real solutions.

Q: How can I solve it by completing the square?

x^2+4x+1=(x+2)^2-3=0. Then (x+2)^2=3, so x=-2\pm\sqrt{3}.

Q: Are the roots rational or irrational?

Since \sqrt{3} is irrational, the solutions x=-2+\sqrt{3} and x=-2-\sqrt{3} are irrational.

Q: What are the sum and product of the roots?

For ax^2+bx+c=0, sum =-\frac{b}{a}=-4 and product =\frac{c}{a}=1. Check: (-2+\sqrt{3})+(-2-\sqrt{3})=-4, product =1.

Answer

We solve the quadratic equation \(x^2+4x+1=0\) using the quadratic formula.

\[
x=\frac{-4\pm\sqrt{4^2-4\cdot 1\cdot 1}}{2\cdot 1}
=\frac{-4\pm\sqrt{16-4}}{2}
=\frac{-4\pm\sqrt{12}}{2}
=\frac{-4\pm 2\sqrt{3}}{2}
=-2\pm\sqrt{3}
\]

Final answer: \(x=-2+\sqrt{3}\) or \(x=-2-\sqrt{3}\).

Detailed Explanation

We want to solve the quadratic equation

\[
x^2+4x+1=0.
\]

This is a quadratic equation of the form

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

so here we identify the coefficients:

\(a=1\)
\(b=4\)
\(c=1\)

We will use the quadratic formula:

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

Now substitute \(a=1\), \(b=4\), and \(c=1\) into the formula:

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

Next, simplify inside the square root:

\[
x=\frac{-4\pm\sqrt{16-4}}{2}.
\]

Compute \(16-4=12\):

\[
x=\frac{-4\pm\sqrt{12}}{2}.
\]

Now simplify \(\sqrt{12}\). Since \(12=4\cdot 3\):

\[
\sqrt{12}=\sqrt{4\cdot 3}= \sqrt{4}\sqrt{3}=2\sqrt{3}.
\]

Substitute this back in:

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

Finally, divide every term in the numerator by \(2\):

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

So the two solutions are:

\[
x_1=-2+\sqrt{3}, \quad x_2=-2-\sqrt{3}.
\]

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Algebra FAQ

How do I solve \(x^2+4x+1=0\) using factoring?

\(\Delta=16-4=12\), so it does not factor over integers. Use the quadratic formula: \(x=\frac{-4\pm\sqrt{12}}{2}=-2\pm\sqrt{3}\).

What is the discriminant and what does it tell me?

\(\Delta=b^2-4ac=4^2-4(1)(1)=12>0\). So there are two distinct real solutions.

What do I get using the quadratic formula?

\(x=\frac{-4\pm\sqrt{4^2-4\cdot1\cdot1}}{2}=\frac{-4\pm\sqrt{12}}{2}=-2\pm\sqrt{3}\).

How can I solve it by completing the square?

\(x^2+4x+1=(x+2)^2-3=0\). Then \((x+2)^2=3\), so \(x=-2\pm\sqrt{3}\).

Are the roots rational or irrational?

Since \(\sqrt{3}\) is irrational, the solutions \(x=-2+\sqrt{3}\) and \(x=-2-\sqrt{3}\) are irrational.

What are the sum and product of the roots?

For \(ax^2+bx+c=0\), sum \(=-\frac{b}{a}=-4\) and product \(=\frac{c}{a}=1\). Check: \((-2+\sqrt{3})+(-2-\sqrt{3})=-4\), product \(=1\).

Solve the quadratic x²+4x+1=0. Use the right method and check.

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