Q. \(x^2 + 2x + 4\)

Q: Can x^2+2x+4 be factored over the integers?

No. Its discriminant is b^2-4ac=2^2-4(1)(4)=4-16=-12<0, so it has no real linear factors. Over integers, it’s irreducible.

Q: What are the roots of x^2+2x+4=0?

Using x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}: x=\frac{-2\pm\sqrt{-12}}{2}=-1\pm i\sqrt{3}.

Q: What is the minimum value of x^2+2x+4 and where does it occur?

Since x^2+2x+4=(x+1)^2+3, the minimum is 3 at x=-1.

Q: What is the vertex and axis of symmetry of y=x^2+2x+4?

Vertex: (-1,3). Axis of symmetry: x=-1.

Q: For what x is x^2+2x+4 equal to a given value k?

Solve x^2+2x+4=k\Rightarrow x^2+2x+(4-k)=0. Discriminant: 4-4(4-k)=4k-12. Hence real solutions when k\ge 3.

Answer

We complete the square:

\[
x^2+2x+4=(x+1)^2+3
\]

So the simplified form is \( (x+1)^2+3 \).

Detailed Explanation

We want to work with the expression

\[
x^2 + 2x + 4
\]

This is already a quadratic expression in standard form \(ax^2 + bx + c\). Here the coefficients are:

\[
a = 1,\quad b = 2,\quad c = 4
\]

Next, we can complete the square to rewrite it in a more informative form.

Step 1: Group the quadratic and linear terms:

\[
x^2 + 2x + 4 = \left(x^2 + 2x\right) + 4
\]

Step 2: Complete the square for \(x^2 + 2x\).

Recall that

\[
x^2 + 2x = (x+1)^2 – 1
\]

Step 3: Substitute this into the original expression:

\[
\left(x^2 + 2x\right) + 4 = \left((x+1)^2 – 1\right) + 4
\]

Step 4: Combine the constants \(-1\) and \(4\):

\[
(x+1)^2 – 1 + 4 = (x+1)^2 + 3
\]

So the expression simplifies to:

\[
x^2 + 2x + 4 = (x+1)^2 + 3
\]

Final answer:

\[
\boxed{x^2 + 2x + 4 = (x+1)^2 + 3}
\]

See full solution

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Algebra FAQ

Can \(x^2+2x+4\) be factored over the integers?

No. Its discriminant is \(b^2-4ac=2^2-4(1)(4)=4-16=-12<0\), so it has no real linear factors. Over integers, it’s irreducible.

What is the completed square form of \(x^2+2x+4\)?

\(x^2+2x+4=(x+1)^2+3\).

What are the roots of \(x^2+2x+4=0\)?

Using \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\): \(x=\frac{-2\pm\sqrt{-12}}{2}=-1\pm i\sqrt{3}\).

What is the minimum value of \(x^2+2x+4\) and where does it occur?

Since \(x^2+2x+4=(x+1)^2+3\), the minimum is \(3\) at \(x=-1\).

What is the vertex and axis of symmetry of \(y=x^2+2x+4\)?

Vertex: \((-1,3)\). Axis of symmetry: \(x=-1\).

For what \(x\) is \(x^2+2x+4\) equal to a given value \(k\)?

Solve \(x^2+2x+4=k\Rightarrow x^2+2x+(4-k)=0\). Discriminant: \(4-4(4-k)=4k-12\). Hence real solutions when \(k\ge 3\).

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