Q. \(x^2+2x=0\)

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

Factor x(x+2)=0. Then x=0 or x=-2.

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

For a=1,b=2,c=0: x=\frac{-2\pm\sqrt{4-0}}{2}=\frac{-2\pm2}{2}. So x=0 or x=-2.

Q: Is there a common factor you can factor out first?

Yes. x^2+2x=x(x+2). Setting each factor to zero gives x=0 or x=-2.

Q: What is the discriminant and what does it tell you?

\Delta=b^2-4ac=2^2-4\cdot1\cdot0=4. Since \Delta>0, there are two real distinct solutions.

Q: Can you solve it by completing the square?

x^2+2x=(x+1)^2-1=0. Then (x+1)^2=1, so x+1=\pm1, giving x=0 or x=-2.

Q: What are the roots and how can you verify them?

Test x=0: 0^2+2\cdot0=0. Test x=-2: (-2)^2+2(-2)=4-4=0.

Answer

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

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

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

Final answers: \(x=0\), \(x=-2\).

Detailed Explanation

We want to solve the equation

\[
x^2+2x=0
\]

Step 1: Factor the left-hand side.

The expression \(x^2+2x\) has a common factor of \(x\). Factor out \(x\):

\[
x^2+2x = x(x+2)
\]

So the equation becomes

\[
x(x+2)=0
\]

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:

\[
x=0 \quad \text{or} \quad x+2=0
\]

Step 3: Solve each simpler equation.

1) From \(x=0\), we get:

\[
x=0
\]

2) From \(x+2=0\), subtract \(2\) from both sides:

\[
x+2-2=0-2
\]
\[
x=-2
\]

Final Answer:

The solutions to \(x^2+2x=0\) are

\[
x=0 \quad \text{and} \quad x=-2
\]

See full solution

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Algebra FAQ

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

Factor \(x(x+2)=0\). Then \(x=0\) or \(x=-2\).

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

For \(a=1,b=2,c=0\): \(x=\frac{-2\pm\sqrt{4-0}}{2}=\frac{-2\pm2}{2}\). So \(x=0\) or \(x=-2\).

Is there a common factor you can factor out first?

Yes. \(x^2+2x=x(x+2)\). Setting each factor to zero gives \(x=0\) or \(x=-2\).

What is the discriminant and what does it tell you?

\(\Delta=b^2-4ac=2^2-4\cdot1\cdot0=4\). Since \(\Delta>0\), there are two real distinct solutions.

Can you solve it by completing the square?

\(x^2+2x=(x+1)^2-1=0\). Then \((x+1)^2=1\), so \(x+1=\pm1\), giving \(x=0\) or \(x=-2\).

What are the roots and how can you verify them?

Test \(x=0\): \(0^2+2\cdot0=0\). Test \(x=-2\): \((-2)^2+2(-2)=4-4=0\).

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