Q. \(x^2 – 5x = 0\)

Q: Solve x^2-5x=0.

Factor: x^2-5x=x(x-5)=0. So x=0 or x=5.

Q: How do you factor x^2-5x?

Take out the common factor x: x^2-5x=x(x-5).

Q: What is the zero-product property here?

If x(x-5)=0, then x=0 or x-5=0, giving x=0 and x=5.

Answer

We solve the equation \(x^2-5x=0\) by factoring:

\[
x^2-5x = x(x-5)=0
\]

So \(x=0\) or \(x-5=0\), giving \(x=5\).

Final result: \(x=0\) or \(x=5\).

Detailed Explanation

We want to solve the equation

\[
x^2 – 5x = 0
\]

Step 1: Factor the left-hand side.

Both terms have a common factor of \(x\), so we factor out \(x\):

\[
x^2 – 5x = x(x – 5)
\]

So the equation becomes

\[
x(x – 5) = 0
\]

Step 2: Use the zero product property.

If a product is zero, then at least one factor must be zero. So we set each factor equal to zero:

\[
x = 0
\]

and

\[
x – 5 = 0
\]

Step 3: Solve each equation.

First solution:

\[
x = 0
\]

Second solution:

Add \(5\) to both sides of \(x – 5 = 0\):

\[
x = 5
\]

Final answer:

\[
x \in \{0, 5\}
\]

See full solution

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Algebra FAQ

Solve \(x^2-5x=0\).

Factor: \(x^2-5x=x(x-5)=0\). So \(x=0\) or \(x=5\).

How do you factor \(x^2-5x\)?

Take out the common factor \(x\): \(x^2-5x=x(x-5)\).

What is the zero-product property here?

If \(x(x-5)=0\), then \(x=0\) or \(x-5=0\), giving \(x=0\) and \(x=5\).

Can you solve it by completing the square?

\(x^2-5x=0\Rightarrow x^2-5x+\left(\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2\). Then \(x=\frac{5}{2}\pm\frac{5}{2}\), so \(x=0,5\).

What are the roots of \(x^2-5x=0\)?

The roots are \(x=0\) and \(x=5\).

What is the factored form of \(x^2-5x\)?

The factored form is \(x(x-5)\).

What is the solution set in interval notation?

The solution set is \(\{0,5\}\).

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