Q. \(x^2+16x+c\)

Q: What is the vertex and minimum value of x^2+16x+c (when a=1)?

Vertex at x=-8. Minimum value is (x+8)^2+(c-64)\Rightarrow c-64.

Q: Solve x^2+16x+c=0 in terms of c.

Using quadratic formula: x=\dfrac{-16\pm\sqrt{256-4c}}{2}=-8\pm\sqrt{64-c}.

Q: When does x^2+16x+c have real roots?

Need discriminant \Delta=256-4c\ge 0\Rightarrow c\le 64.

Q: For which c is there exactly one real root (tangent case)?

Exactly one root when \Delta=0\Rightarrow 256-4c=0\Rightarrow c=64.

Q: For which c is x^2+16x+c always positive?

Since minimum is c-64, require c-64>0\Rightarrow c>64.

Q: Expand (x+8)^2+(c-64) to verify equivalence.

(x+8)^2+(c-64)=x^2+16x+64+c-64=x^2+16x+c.

Answer

\(x^2+16x+c\) is a quadratic in \(x\). To factor it, find two numbers that multiply to \(c\) and add to \(16\). So it factors as

\[
x^2+16x+c=(x+a)(x+b)
\]
where \(a+b=16\) and \(ab=c\). Equivalently, the fully completed-square form is

\[
x^2+16x+c=(x+8)^2+(c-64).
\]

Detailed Explanation

We are asked to work with the expression

\[
\text{Let } E(x)=x^2+16x+c.
\]

Here, \(c\) is a constant. A common next step for a quadratic like \(x^2+16x+c\) is to rewrite it in vertex (completed square) form.

Step 1: Focus on the variable terms \(x^2+16x\).

\[
x^2+16x = x^2+2\cdot x \cdot 8.
\]

Step 2: Complete the square.

To complete the square, we add \((8)^2\) inside the expression:

\[
x^2+16x = \left(x^2+16x+64\right)-64.
\]

Step 3: Recognize the perfect square trinomial.

\[
x^2+16x+64 = (x+8)^2.
\]

So we have:

\[
x^2+16x = (x+8)^2-64.
\]

Step 4: Substitute back into the full expression \(x^2+16x+c\).

\[
x^2+16x+c = \left((x+8)^2-64\right)+c.
\]

Step 5: Combine constants.

\[
-64+c = c-64.
\]

So the expression becomes:

\[
x^2+16x+c = (x+8)^2 + (c-64).
\]

Final Answer:

\[
x^2+16x+c = (x+8)^2 + c – 64.
\]

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Algebra FAQ

Complete the square for \(x^2+16x+c\).

\(x^2+16x+c=(x+8)^2+(c-64)\).

What is the vertex and minimum value of \(x^2+16x+c\) (when \(a=1\))?

Vertex at \(x=-8\). Minimum value is \((x+8)^2+(c-64)\Rightarrow c-64\).

Solve \(x^2+16x+c=0\) in terms of \(c\).

Using quadratic formula: \(x=\dfrac{-16\pm\sqrt{256-4c}}{2}=-8\pm\sqrt{64-c}\).

When does \(x^2+16x+c\) have real roots?

Need discriminant \(\Delta=256-4c\ge 0\Rightarrow c\le 64\).

For which \(c\) is there exactly one real root (tangent case)?

Exactly one root when \(\Delta=0\Rightarrow 256-4c=0\Rightarrow c=64\).

For which \(c\) is \(x^2+16x+c\) always positive?

Since minimum is \(c-64\), require \(c-64>0\Rightarrow c>64\).

Expand \((x+8)^2+(c-64)\) to verify equivalence.

\((x+8)^2+(c-64)=x^2+16x+64+c-64=x^2+16x+c\).

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