Solve Quadratic Equation: x² - 4x + 4 = 0 Step-by-Step

Q: How do I recognize a perfect square trinomial?

Look for the pattern $ a^2 - 2ab + b^2 $! In $ x^2 - 4x + 4 $, we have x² (first square), -4x (twice the product), and 4 (second square of 2).

Q: Why does this equation have only one solution?

When $ (x-2)^2 = 0 $, the factor $ (x-2) $ appears twice. This means x = 2 is a repeated root - it's counted twice but has the same value!

Q: Can I solve this using the quadratic formula instead?

Yes, but it's unnecessary! The quadratic formula gives $ x = \frac{4 \pm \sqrt{16-16}}{2} = \frac{4}{2} = 2 $. Factoring is much faster for perfect squares.

Q: What if I expand (x-2)² and don't get the original equation?

Double-check your expansion: $ (x-2)^2 = (x-2)(x-2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4 $. Remember the middle term comes from adding both cross-products!

Q: How do I know when to look for perfect square patterns?

The first and last terms are perfect squaresThe middle term equals twice the product of the square rootsAll signs follow the $ (a \pm b)^2 $ pattern

Quadratic Equations with Perfect Square Trinomials

Solve the following equation:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Use the roots formula

00:25 Identify the coefficients

00:39 Substitute appropriate values according to the given data and solve

01:09 Calculate the square and products

01:28 The square root of 0 equals 0

01:35 When the root equals 0, there is only one solution to the equation

01:47 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equation:

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

Step-by-step solution

The given equation is:

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

This resembles a perfect square trinomial. The expression $x^2 - 4x + 4$ can be rewritten as $(x-2)^2$ . This can be verified by expanding $(x-2)(x-2)$ to confirm it equals $x^2 - 4x + 4$ .

Therefore, the equation becomes:

$(x-2)^2 = 0$

To solve for $x$ , take the square root of both sides:

$x - 2 = 0$

Adding 2 to both sides gives:

$x = 2$

Thus, the solution to the equation $x^2 - 4x + 4 = 0$ is $x = 2$ , which corresponds to the unique real root of the equation.

Final Answer

$x=2$

Key Points to Remember

Essential concepts to master this topic

Pattern Recognition: Identify $x^2 - 4x + 4$ as perfect square trinomial
Factoring Method: Rewrite as $(x-2)^2 = 0$ using square pattern
Solution Check: Substitute x = 2: $2^2 - 4(2) + 4 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Using quadratic formula when factoring is simpler
Don't automatically apply the quadratic formula to $x^2 - 4x + 4 = 0$ = messy calculations! This wastes time and increases error risk. Always check first if the trinomial is a perfect square that factors easily as $(x-2)^2$ .

Practice Quiz

Test your knowledge with interactive questions

a = Coefficient of x²

b = Coefficient of x

c = Coefficient of the independent number

what is the value of $$ a $$ in the equation

$$ y=3x-10+5x^2 $$

FAQ

Everything you need to know about this question

How do I recognize a perfect square trinomial?

Look for the pattern $a^2 - 2ab + b^2$ ! In $x^2 - 4x + 4$ , we have x² (first square), -4x (twice the product), and 4 (second square of 2).

Why does this equation have only one solution?

When $(x-2)^2 = 0$ , the factor $(x-2)$ appears twice. This means x = 2 is a repeated root - it's counted twice but has the same value!

Can I solve this using the quadratic formula instead?

Yes, but it's unnecessary! The quadratic formula gives $x = \frac{4 \pm \sqrt{16-16}}{2} = \frac{4}{2} = 2$ . Factoring is much faster for perfect squares.

What if I expand (x-2)² and don't get the original equation?

Double-check your expansion: $(x-2)^2 = (x-2)(x-2) = x^2 - 2x - 2x + 4 = x^2 - 4x + 4$ . Remember the middle term comes from adding both cross-products!

How do I know when to look for perfect square patterns?

The first and last terms are perfect squares
The middle term equals twice the product of the square roots
All signs follow the $(a \pm b)^2$ pattern

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