Q. \(x^2+5x=0\)

Q: How do I factor x^2+5x=0?

Factor out x: \,x(x+5)=0. Then set each factor to zero: x=0 or x+5=0.

Q: Can I solve x^2+5x=0 using the quadratic formula?

With a=1,b=5,c=0: x=\frac{-5\pm\sqrt{25}}{2}=\frac{-5\pm5}{2}, so x=0 or x=-5.

Q: What is the role of the common factor in x^2+5x?

Both terms share x, so x^2+5x=x(x+5). Factoring is faster than expanding or using more complex methods.

Q: Why is there no x-constant term here?

Because c=0 in ax^2+bx+c. That guarantees one root is x=0 since the equation becomes x(\cdots)=0.

Q: How can I check the solutions quickly?

Substitute: for x=0, 0^2+5\cdot0=0. For x=-5, (-5)^2+5(-5)=25-25=0.

Answer

We solve \(x^2+5x=0\) by factoring:

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

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

Final answers: \(x=0\) and \(x=-5\).

Detailed Explanation

We want to solve the equation:

\[
x^2 + 5x = 0
\]

Step 1: Factor out the common factor.

Both terms contain \(x\), so we factor \(x\) out:

\[
x(x + 5) = 0
\]

Step 2: Use the zero product property.

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

\[
x = 0 \quad \text{or} \quad x + 5 = 0
\]

Step 3: Solve the first equation.

If \(x = 0\), then one solution is:

\[
x = 0
\]

Step 4: Solve the second equation.

Now solve \(x + 5 = 0\):

\[
x + 5 = 0
\]

Subtract \(5\) from both sides:

\[
x = -5
\]

Final Answer:

The solutions to \(x^2 + 5x = 0\) are:

\[
x = 0 \quad \text{and} \quad x = -5
\]

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Algebra FAQ

How do I factor \(x^2+5x=0\)?

Factor out \(x\): \(\,x(x+5)=0\). Then set each factor to zero: \(x=0\) or \(x+5=0\).

What are the solutions to \(x^2+5x=0\)?

\(\,x(x+5)=0\) gives \(x=0\) and \(x=-5\).

Can I solve \(x^2+5x=0\) using the quadratic formula?

With \(a=1,b=5,c=0\): \(x=\frac{-5\pm\sqrt{25}}{2}=\frac{-5\pm5}{2}\), so \(x=0\) or \(x=-5\).

What is the role of the common factor in \(x^2+5x\)?

Both terms share \(x\), so \(x^2+5x=x(x+5)\). Factoring is faster than expanding or using more complex methods.

Why is there no \(x\)-constant term here?

Because \(c=0\) in \(ax^2+bx+c\). That guarantees one root is \(x=0\) since the equation becomes \(x(\cdots)=0\).

How can I check the solutions quickly?

Substitute: for \(x=0\), \(0^2+5\cdot0=0\). For \(x=-5\), \((-5)^2+5(-5)=25-25=0\).

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