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

Perfect Square Trinomials with Double Roots

Solve the following equation:

x24x+4=0 x^2-4x+4=0

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:06 Let's find the value of X.
00:10 First, identify the coefficients in the equation.
00:25 Now, use the formula for finding the roots of the equation.
00:49 Substitute the given values into the formula and solve it step by step.
01:13 Next, calculate the square and multiply the products.
01:33 Remember, the square root of zero is zero.
01:40 When the root is zero, the equation has only one solution.
01:56 And that's how we find the solution to this problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

x24x+4=0 x^2-4x+4=0

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Recognize the given quadratic equation x24x+4=0 x^2 - 4x + 4 = 0 as a perfect square trinomial.
  • Step 2: Identify the factored form, which is (x2)2=0 (x - 2)^2 = 0 .
  • Step 3: Solve the factored equation to find the value of x x .

Now, let's work through each step:

Step 1: The problem gives us the quadratic equation x24x+4=0 x^2 - 4x + 4 = 0 . This equation is a perfect square trinomial because it can be rewritten as (x2)2=0 (x - 2)^2 = 0 .

Step 2: Recognize and rewrite the equation in its factored form:

(x2)2=0 (x - 2)^2 = 0 .

Step 3: To solve this factored equation, set the factor equal to zero:

x2=0 x - 2 = 0 .

Solving for x x , we get:

x=2 x = 2 .

In this case, the equation has a double root, x=2 x = 2 .

Therefore, the solution to the problem is x=2 x = 2 .

3

Final Answer

x=2 x=2

Key Points to Remember

Essential concepts to master this topic
  • Perfect Square: Recognize x24x+4=(x2)2 x^2 - 4x + 4 = (x-2)^2 pattern
  • Factoring: Check middle term: 4x=22x -4x = -2 \cdot 2 \cdot x
  • Verify: Substitute x=2 x = 2 : 48+4=0 4 - 8 + 4 = 0

Common Mistakes

Avoid these frequent errors
  • Finding two different roots for perfect square trinomials
    Don't assume all quadratics have two different solutions like x1=2,x2=2 x_1 = 2, x_2 = -2 ! Perfect square trinomials have repeated roots, meaning the same solution appears twice. Always recognize the pattern (xa)2=0 (x-a)^2 = 0 gives only one unique solution.

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

Why does this quadratic have only one solution instead of two?

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Great question! This equation has a double root or repeated root. When you factor x24x+4=(x2)2 x^2 - 4x + 4 = (x-2)^2 , you get the same factor twice. So technically there are two roots, but they're both equal to 2!

How can I quickly recognize a perfect square trinomial?

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Look for the pattern a22ab+b2 a^2 - 2ab + b^2 ! In x24x+4 x^2 - 4x + 4 :

  • First term: x2=(x)2 x^2 = (x)^2
  • Last term: 4=(2)2 4 = (2)^2
  • Middle term: 4x=2x2 -4x = -2 \cdot x \cdot 2

If all three match this pattern, it's a perfect square!

What if I can't see the perfect square pattern immediately?

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No problem! You can always use the quadratic formula or try to factor by grouping. However, learning to spot perfect squares saves time and reduces calculation errors.

Are there other ways to solve this equation?

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Yes! You could use:

  • Quadratic formula: x=4±16162=4±02=2 x = \frac{4 \pm \sqrt{16-16}}{2} = \frac{4 \pm 0}{2} = 2
  • Completing the square: Though it's already a perfect square!
  • Graphing: Find where the parabola touches the x-axis

How do I check my answer when there's only one root?

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Substitute x=2 x = 2 into the original equation:

(2)24(2)+4=48+4=0 (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0

Since you get 0, your answer is correct!

What does a double root mean graphically?

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Graphically, a double root means the parabola touches the x-axis at exactly one point (the vertex) but doesn't cross it. At x=2 x = 2 , the graph just touches the x-axis and bounces back up!

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