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

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

Factor: \[x^2 - x = x(x-1)=0.\] Then x=0 or x=1.

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

The roots are the solutions to x(x-1)=0, so x=0 and x=1.

Q: Can you solve using the quadratic formula?

Rewrite as x^2 - x + 0=0. Then \[x=\frac{1\pm\sqrt{(-1)^2-4\cdot1\cdot0}}{2}=\frac{1\pm1}{2},\] giving x=0,1.

Q: Is there a common factor method for x^2-x=0?

Yes, factor out x: \[x^2-x=x(x-1).\] Set each factor to zero: x=0 or x-1=0\Rightarrow x=1.

Q: What is the factored form of x^2-x=0?

\[x^2-x = x(x-1).\] The equation becomes x(x-1)=0.

Answer

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

\[x^2 – x = x(x-1) = 0\]

So \(x=0\) or \(x-1=0\), which gives \(x=1\).

Final answers: \(x=0\) and \(x=1\).

Detailed Explanation

We want to solve the equation

\[
x^2 – x = 0.
\]

Step 1: Factor the left-hand side.

The expression \(x^2 – x\) is a difference of terms that can be factored by taking out \(x\):

\[
x^2 – x = x(x – 1).
\]

Step 2: Use the zero-product property.

The equation becomes

\[
x(x – 1) = 0.
\]

The zero-product property says: if

\[
ab = 0,
\]

then either \(a = 0\) or \(b = 0\).

Step 3: Set each factor equal to zero.

First factor:

\[
x = 0.
\]

Second factor:

\[
x – 1 = 0.
\]

Step 4: Solve the second equation.

Add \(1\) to both sides of \(x – 1 = 0\):

\[
x = 1.
\]

Final Answer.

The solutions are

\[
x = 0 \quad \text{or} \quad x = 1.
\]

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Algebra FAQ

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

Factor: \[x^2 - x = x(x-1)=0.\] Then \(x=0\) or \(x=1\).

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

The roots are the solutions to \(x(x-1)=0\), so \(x=0\) and \(x=1\).

Can you solve using the quadratic formula?

Rewrite as \(x^2 - x + 0=0.\) Then \[x=\frac{1\pm\sqrt{(-1)^2-4\cdot1\cdot0}}{2}=\frac{1\pm1}{2},\] giving \(x=0,1\).

Is there a common factor method for \(x^2-x=0\)?

Yes, factor out \(x\): \[x^2-x=x(x-1).\] Set each factor to zero: \(x=0\) or \(x-1=0\Rightarrow x=1\).

What is the factored form of \(x^2-x=0\)?

\[x^2-x = x(x-1).\] The equation becomes \(x(x-1)=0\).

How many solutions does \(x^2-x=0\) have?

There are two real solutions because it factors into two distinct linear factors: \(x=0\) and \(x=1\).

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