Q. \(x^{2}-4=0\)

Q: Solve x^2-4=0.

Factor: x^2-4=(x-2)(x+2)=0. So x=2 or x=-2.

Q: What are the roots of x^2-4=0 using square roots?

x^2=4. Then x=\pm\sqrt{4}=\pm2, giving x=2 and x=-2.

Q: Can I use the quadratic formula on x^2-4=0?

For ax^2+bx+c=0, here a=1,b=0,c=-4. x=\frac{-0\pm\sqrt{0-4(1)(-4)}}{2(1)}=\frac{\pm\sqrt{16}}{2}=\pm2.

Q: How do you factor x^2-4 correctly?

Recognize difference of squares: x^2-4=x^2-2^2=(x-2)(x+2).

Q: What is the discriminant and how does it affect solutions?

Discriminant b^2-4ac=0^2-4(1)(-4)=16. Since it is positive, there are two real solutions: x=\pm2.

Q: Why are there two solutions instead of one?

x^2=4 means both positive and negative numbers square to 4. Thus x=2 and x=-2.

Answer

We solve the equation \(x^2-4=0\) by factoring:

\[
x^2-4=(x-2)(x+2)=0
\]

So \(x-2=0\) or \(x+2=0\), giving \(x=2\) or \(x=-2\).

Final result: \(x=2\) or \(x=-2\).

Detailed Explanation

Problem: Solve the equation \(x^2 – 4 = 0\).

Step 1: Add 4 to both sides

We want to isolate the \(x^2\) term. Starting with

\[
x^2 – 4 = 0
\]

Add \(4\) to both sides:

\[
x^2 – 4 + 4 = 0 + 4
\]

This simplifies to

\[
x^2 = 4
\]

Step 2: Take the square root of both sides

Now we solve \(x^2 = 4\) by taking the square root on both sides. Since squaring loses the sign information, we must consider both positive and negative square roots:

\[
x = \pm \sqrt{4}
\]

Compute the square root:

\[
\sqrt{4} = 2
\]

Step 3: Write both solutions

Therefore,

\[
x = 2 \quad \text{or} \quad x = -2
\]

Answer: The solutions to \(x^2 – 4 = 0\) are \(x = 2\) and \(x = -2\).

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Algebra FAQ

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

Factor: \(x^2-4=(x-2)(x+2)=0\). So \(x=2\) or \(x=-2\).

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

\(x^2=4\). Then \(x=\pm\sqrt{4}=\pm2\), giving \(x=2\) and \(x=-2\).

Can I use the quadratic formula on \(x^2-4=0\)?

For \(ax^2+bx+c=0\), here \(a=1,b=0,c=-4\). \(x=\frac{-0\pm\sqrt{0-4(1)(-4)}}{2(1)}=\frac{\pm\sqrt{16}}{2}=\pm2\).

How do you factor \(x^2-4\) correctly?

Recognize difference of squares: \(x^2-4=x^2-2^2=(x-2)(x+2)\).

What is the discriminant and how does it affect solutions?

Discriminant \(b^2-4ac=0^2-4(1)(-4)=16\). Since it is positive, there are two real solutions: \(x=\pm2\).

Why are there two solutions instead of one?

\(x^2=4\) means both positive and negative numbers square to \(4\). Thus \(x=2\) and \(x=-2\).

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