Q. \(x^2 – 16 = 0\)

Q: How do you factor x^2-16?

Use difference of squares: x^2-16=x^2-4^2=(x-4)(x+4).

Q: What are the solutions from the factored form?

From (x-4)(x+4)=0, set x-4=0 or x+4=0. So x=4 or x=-4.

Q: How do you use square roots to solve it?

x^2=16. Take square roots: x=\pm\sqrt{16}=\pm 4.

Q: What is the discriminant method here?

For x^2+0x-16=0, a=1,b=0,c=-16. Discriminant D=b^2-4ac=0-4(1)(-16)=64. x=\frac{-b\pm\sqrt{D}}{2a}=\pm 4.

Q: Check the solutions in the original equation.

If x=4: 4^2-16=16-16=0. If x=-4: (-4)^2-16=16-16=0. Both work.

Q: What is the general rule for equations x^2-a^2=0?

x^2-a^2=0\Rightarrow x^2=a^2\Rightarrow x=\pm a. Here a=4, so x=\pm 4.

Answer

We solve the quadratic equation \(x^2 – 16 = 0\).

Add 16 to both sides:

\[
x^2 = 16
\]

Take the square root of both sides:

\[
x = \pm 4
\]

Final result: \(x = 4\) or \(x = -4\).

Detailed Explanation

We want to solve the equation \(x^2 – 16 = 0\).

Step 1: Add \(16\) to both sides

Start with:

\[
x^2 – 16 = 0
\]

Add \(16\) to both sides so that the \(-16\) is eliminated from the left side.

\[
x^2 – 16 + 16 = 0 + 16
\]

This simplifies to:

\[
x^2 = 16
\]

Step 2: Take the square root of both sides

Now solve for \(x\) by taking square roots. Since \(x^2 = 16\), \(x\) can be both the positive and negative square roots of \(16\).

\[
x = \pm \sqrt{16}
\]

Compute \(\sqrt{16}\):

\[
\sqrt{16} = 4
\]

So:

\[
x = \pm 4
\]

Final Answer

The solutions are:

\[
x = 4 \quad \text{or} \quad x = -4
\]

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Algebra FAQ

Solve \(x^2-16=0\).

\(x^2=16\Rightarrow x=\pm 4\).

How do you factor \(x^2-16\)?

Use difference of squares: \(x^2-16=x^2-4^2=(x-4)(x+4)\).

What are the solutions from the factored form?

From \((x-4)(x+4)=0\), set \(x-4=0\) or \(x+4=0\). So \(x=4\) or \(x=-4\).

How do you use square roots to solve it?

\(x^2=16\). Take square roots: \(x=\pm\sqrt{16}=\pm 4\).

What is the discriminant method here?

For \(x^2+0x-16=0\), \(a=1,b=0,c=-16\). Discriminant \(D=b^2-4ac=0-4(1)(-16)=64\). \(x=\frac{-b\pm\sqrt{D}}{2a}=\pm 4\).

Check the solutions in the original equation.

If \(x=4\): \(4^2-16=16-16=0\). If \(x=-4\): \((-4)^2-16=16-16=0\). Both work.

What is the general rule for equations \(x^2-a^2=0\)?

\(x^2-a^2=0\Rightarrow x^2=a^2\Rightarrow x=\pm a\). Here \(a=4\), so \(x=\pm 4\).

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