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

Q: How do you solve x^2-4x=0 by factoring?

Factor out x: x(x-4)=0. So x=0 or x=4.

Q: How do you solve x^2-4x=0 using the quadratic formula?

For ax^2+bx+c=0, a=1, b=-4, c=0. Then x=\frac{4\pm\sqrt{16}}{2}=\frac{4\pm 4}{2}, giving x=0 or x=4.

Q: What is the factoring method when there is a common factor like x?

Identify the GCF: x^2-4x=x(x-4). Use the zero product property: set each factor to zero.

Q: How can you check your solutions for x^2-4x=0?

Substitute x=0: 0-0=0. Substitute x=4: 16-16=0. Both satisfy the equation.

Q: What is the zero product property used here?

If x(x-4)=0, then x=0 or x-4=0. This works because a product is zero only when at least one factor is zero.

Q: What are the roots and multiplicities of x^2-4x=0?

The roots are x=0 and x=4. Each factor is linear and appears once, so both roots have multiplicity 1.

Answer

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

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

So \(x=0\) or \(x=4\).

Final answer: \(x=0\) or \(x=4\).

Detailed Explanation

We want to solve the equation:

\[
x^2 – 4x = 0
\]

Step 1: Factor the left-hand side.

Both terms on the left contain a factor of \(x\). Factor out \(x\):

\[
x^2 – 4x = x(x – 4)
\]

So the equation becomes:

\[
x(x – 4) = 0
\]

Step 2: Use the zero product property.

The zero product property says: if

\[
A \cdot B = 0
\]

then

\[
A = 0 \quad \text{or} \quad B = 0
\]

Here, \(A = x\) and \(B = x – 4\). So we set each factor equal to zero.

Step 3: Solve \(x = 0\).

One possibility is:

\[
x = 0
\]

Step 4: Solve \(x – 4 = 0\).

The other possibility is:

\[
x – 4 = 0
\]

Add \(4\) to both sides:

\[
x = 4
\]

Final Answer:

The solutions are:

\[
x = 0 \quad \text{or} \quad x = 4
\]

See full solution

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Algebra FAQ

How do you solve \(x^2-4x=0\) by factoring?

Factor out \(x\): \(x(x-4)=0\). So \(x=0\) or \(x=4\).

How do you solve \(x^2-4x=0\) using the quadratic formula?

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

What is the factoring method when there is a common factor like \(x\)?

Identify the GCF: \(x^2-4x=x(x-4)\). Use the zero product property: set each factor to zero.

How can you check your solutions for \(x^2-4x=0\)?

Substitute \(x=0\): \(0-0=0\). Substitute \(x=4\): \(16-16=0\). Both satisfy the equation.

What is the zero product property used here?

If \(x(x-4)=0\), then \(x=0\) or \(x-4=0\). This works because a product is zero only when at least one factor is zero.

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

The roots are \(x=0\) and \(x=4\). Each factor is linear and appears once, so both roots have multiplicity \(1\).

Solve: find the roots of x²-4x=0.
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