x^2>4 implies x 2. In interval notation: (-\infty,-2)\cup(2,\infty).

Q. \[ x^2 – 4 \]

Q: What is the vertex form of x^2-4?

It is x^2-4=(x-0)^2-4, so the vertex is (0,-4).

Q: Does x^2-4 have a real y-intercept?

Yes. At x=0, y=0^2-4=-4, so intercept is (0,-4).

Q: What are the sign intervals of x^2-4?

Using factor form: x^2-4>0 for x 2; x^2-4<0 for -2<x<2. Zero at x=\pm 2.

Q: What is the definite integral \int (x^2-4)\,dx?

\int (x^2-4)\,dx=\frac{x^3}{3}-4x+C.

Answer

We factor the expression \(x^2-4\) as a difference of squares.

\[
x^2-4=(x-2)(x+2)
\]

So the solutions to \(x^2-4=0\) are \(x=2\) and \(x=-2\).

Detailed Explanation

We are given the expression \(x^2 – 4\). Our goal is usually to simplify it by factoring it (because \(x^2 – 4\) is a difference of squares).

Step 1: Recognize a difference of squares

Recall the identity:

\[
a^2 – b^2 = (a-b)(a+b)
\]

Rewrite \(x^2 – 4\) in the form \(a^2 – b^2\).

Match terms:

\(a^2 = x^2\) so \(a = x\).
\(b^2 = 4\) so \(b = 2\) (since \(2^2 = 4\)).

Step 2: Apply the identity

Substitute \(a = x\) and \(b = 2\) into the formula \(a^2 – b^2 = (a-b)(a+b)\).

\[
x^2 – 4 = (x-2)(x+2)
\]

Final Answer

\[
x^2 – 4 = (x-2)(x+2)
\]

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Algebra FAQ

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

Solve \(x^2=4\), so \(x=\pm 2\).

Can \(x^2-4\) be factored?

Yes. \(x^2-4=(x-2)(x+2)\).

What is the vertex form of \(x^2-4\)?

It is \(x^2-4=(x-0)^2-4\), so the vertex is \((0,-4)\).

Does \(x^2-4\) have a real \(y\)-intercept?

Yes. At \(x=0\), \(y=0^2-4=-4\), so intercept is \((0,-4)\).

What are the sign intervals of \(x^2-4\)?

Using factor form: \(x^2-4>0\) for \(x<-2\) or \(x>2\); \(x^2-4<0\) for \(-2

Solve \(x^2-4>0\).

\(x^2>4\) implies \(x<-2\) or \(x>2\). In interval notation: \((-\infty,-2)\cup(2,\infty)\).

What is the derivative of \(x^2-4\)?

\(\frac{d}{dx}(x^2-4)=2x\).

What is the definite integral \(\int (x^2-4)\,dx\)?

\(\int (x^2-4)\,dx=\frac{x^3}{3}-4x+C\).

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