Q. \(x^2 + 12x + 36 = 0\)

Q: How do I factor x^2+12x+36=0 quickly?

x^2+12x+36=(x+6)^2. So (x+6)^2=0 and x=-6.

Q: What are the roots using the quadratic formula?

For ax^2+bx+c=0, a=1,b=12,c=36. Discriminant D=12^2-4\cdot1\cdot36=0. Thus x=\frac{-12\pm0}{2}=-6.

Q: What is the discriminant and what does it mean here?

D=b^2-4ac=144-144=0. This means one real repeated root.

Q: Can I solve by completing the square?

x^2+12x+36=0\Rightarrow x^2+12x=-36. Add (\frac{12}{2})^2=36: x^2+12x+36=(x+6)^2=0. Hence x=-6.

Q: Why is there only one solution?

Because the quadratic is a perfect square: (x+6)^2=0. A squared expression equals zero only when x+6=0, giving the single root x=-6.

Q: What is the vertex of the parabola and where is it relative to the x-axis?

For x^2+12x+36, vertex at x=-\frac{b}{2a}=-6. Value there is 0, so it touches the x-axis at x=-6.

Answer

We solve the quadratic \(x^2 + 12x + 36 = 0\) by factoring.

\(x^2 + 12x + 36 = (x+6)^2\)

So \((x+6)^2 = 0 \Rightarrow x = -6\).

Final result: \(x=-6\).

Detailed Explanation

We need to solve the quadratic equation

\[
x^2+12x+36=0
\]

Step 1: Identify a factoring pattern

For a quadratic equation of the form

\[
x^2+bx+c=0,
\]

we look for two numbers that multiply to \(c\) and add to \(b\).

Here, \(b=12\) and \(c=36\). We look for numbers \(m\) and \(n\) such that

\[
m\cdot n=36
\]

and

\[
m+n=12.
\]

Step 2: Find the correct numbers

The numbers \(6\) and \(6\) work because

\[
6\cdot 6=36
\]

and

\[
6+6=12.
\]

Step 3: Factor the quadratic

Now factor using \((x+m)(x+n)\):

\[
x^2+12x+36=(x+6)(x+6).
\]

So the equation becomes

\[
(x+6)(x+6)=0.
\]

Step 4: Use the zero product property

\[
(x+6)(x+6)=0,
\]

then at least one factor must be zero. So we set

\[
x+6=0.
\]

Step 5: Solve for \(x\)

\[
x+6=0 \quad \Rightarrow \quad x=-6.
\]

Final Answer

The solution is

\[
x=-6.
\]

See full solution

Graph

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Algebra FAQ

How do I factor \(x^2+12x+36=0\) quickly?

\(x^2+12x+36=(x+6)^2\). So \((x+6)^2=0\) and \(x=-6\).

What are the roots using the quadratic formula?

For \(ax^2+bx+c=0\), \(a=1,b=12,c=36\). Discriminant \(D=12^2-4\cdot1\cdot36=0\). Thus \(x=\frac{-12\pm0}{2}=-6\).

What is the discriminant and what does it mean here?

\(D=b^2-4ac=144-144=0\). This means one real repeated root.

Can I solve by completing the square?

\(x^2+12x+36=0\Rightarrow x^2+12x=-36\). Add \((\frac{12}{2})^2=36\): \(x^2+12x+36=(x+6)^2=0\). Hence \(x=-6\).

Why is there only one solution?

Because the quadratic is a perfect square: \((x+6)^2=0\). A squared expression equals zero only when \(x+6=0\), giving the single root \(x=-6\).

What is the vertex of the parabola and where is it relative to the x-axis?

For \(x^2+12x+36\), vertex at \(x=-\frac{b}{2a}=-6\). Value there is \(0\), so it touches the x-axis at \(x=-6\).

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