Q. \(x^{2}+4x+4=0\)

Q: Find the solutions of x^2+4x+4=0.

x^2+4x+4=(x+2)^2. So (x+2)^2=0\Rightarrow x=-2 (double root).

Q: How do you factor x^2+4x+4.

It is a perfect square: x^2+4x+4=(x+2)^2. So factoring gives (x+2)(x+2).

Q: Solve using the quadratic formula.

For ax^2+bx+c=0, a=1, b=4, c=4. Then x=\frac{-4\pm\sqrt{16-16}}{2}=\frac{-4}{2}=-2.

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

\Delta=b^2-4ac=16-16=0. Discriminant 0 means one real repeated root (multiplicity 2).

Q: Complete the square for x^2+4x+4=0.

x^2+4x+4=(x+2)^2. Set (x+2)^2=0, giving x=-2.

Q: Check the solution by substitution.

Substitute x=-2: (-2)^2+4(-2)+4=4-8+4=0. So it satisfies the equation.

Q: Is there any other solution besides x=-2.

No. Since (x+2)^2=0, the only solution is x=-2, repeated twice.

Answer

We solve the quadratic \(x^2+4x+4=0\) by factoring:

\[
x^2+4x+4=(x+2)(x+2)=(x+2)^2
\]

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

\(x=-2\) (double root).

Detailed Explanation

We want to solve the equation

\[
x^2+4x+4=0.
\]

Step 1: Recognize a perfect square.

Notice that

\[
(x+2)^2 = x^2+4x+4.
\]

So the left side of the equation matches \( (x+2)^2 \).

Step 2: Rewrite the equation using the perfect square.

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

Step 3: Use the zero product idea for a square.

If \( (x+2)^2=0 \), then the only way a square equals \(0\) is if the expression inside the square equals \(0\).

\[
x+2=0.
\]

Step 4: Solve the linear equation for \(x\).

\[
x=-2.
\]

Final Answer:

\[
x=-2.
\]

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Algebra FAQ

Find the solutions of \(x^2+4x+4=0\).

\(x^2+4x+4=(x+2)^2\). So \((x+2)^2=0\Rightarrow x=-2\) (double root).

How do you factor \(x^2+4x+4\).

It is a perfect square: \(x^2+4x+4=(x+2)^2\). So factoring gives \((x+2)(x+2)\).

Solve using the quadratic formula.

For \(ax^2+bx+c=0\), \(a=1\), \(b=4\), \(c=4\). Then \(x=\frac{-4\pm\sqrt{16-16}}{2}=\frac{-4}{2}=-2\).

What is the discriminant and what does it mean here?

\( \Delta=b^2-4ac=16-16=0\). Discriminant \(0\) means one real repeated root (multiplicity 2).

Complete the square for \(x^2+4x+4=0\).

\(x^2+4x+4=(x+2)^2\). Set \((x+2)^2=0\), giving \(x=-2\).

Check the solution by substitution.

Substitute \(x=-2\): \((-2)^2+4(-2)+4=4-8+4=0\). So it satisfies the equation.

Is there any other solution besides \(x=-2\).

No. Since \((x+2)^2=0\), the only solution is \(x=-2\), repeated twice.

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