To solve the quadratic equation $x^2 - 4x + 4 = 0$ using the quadratic formula, follow these steps:

Step 1: Identify the coefficients

In the quadratic equation $ax^2 + bx + c = 0$, identify the values of $a$, $b$, and $c$.

For $x^2 - 4x + 4 = 0$:

• $a = 1$
• $b = -4$
• $c = 4$

Step 2: Write the quadratic formula

The quadratic formula is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Step 3: Substitute the values of $a$, $b$, and $c$

Substitute $a = 1$, $b = -4$, and $c = 4$ into the quadratic formula:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(4)}}{2(1)}$

Step 4: Simplify inside the square root

First, calculate $b^2 - 4ac$:

$(-4)^2 - 4(1)(4) = 16 - 16 = 0$

So the expression simplifies to:

$x = \frac{4 \pm \sqrt{0}}{2}$

Step 5: Simplify further

Since $\sqrt{0} = 0$, the quadratic formula becomes:

$x = \frac{4 \pm 0}{2}$

$x = \frac{4}{2}$

$x = 2$

Step 6: Final solution

There is only one solution: $x = 2$.

This quadratic equation has a double root, meaning the solution $x = 2$ has multiplicity 2.

---

Would you like more details, or do you have any questions?

5 Related Questions:

1. How do you recognize when a quadratic equation has a double root?
2. How do you factor a quadratic equation like this without using the quadratic formula?
3. What are the conditions for a quadratic equation to have real, distinct roots?
4. How does the discriminant $b^2 - 4ac$ affect the nature of the solutions?
5. How would the solution change if the constant term were different?

Tip:

The discriminant ($b^2 - 4ac$) determines the type of roots a quadratic equation will have: if it's zero, there's one real double root; if positive, two distinct real roots; and if negative, two complex roots.