Solve the Perfect Square Equation: (x-5)² = 0

Perfect Square Equations with Zero Property

Find the value of the parameter x.

(x−5)2=0 (x-5)^2=0

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Step-by-step video solution

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00:00 Solve
00:03 Extract root
00:06 Square root cancels square
00:10 Isolate the unknown
00:13 This is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find the value of the parameter x.

(x−5)2=0 (x-5)^2=0

2

Step-by-step solution

We will factor using the shortened multiplication formulas:

a2−b2=(a−b)(a+b) a^2-b^2=(a-b)(a+b) (a−b)2=a2−2ab+b2 (a-b)^2=a^2-2ab+b^2

(a+b)2=a2+2ab+b2 (a+b)^2=a^2+2ab+b^2

Let's remember that there might be more than one solution for the value of x.

According to one solution, we'll take the square root:

(x−5)2=0 (x-5)^2=0

x−5=0 x-5=0

x=5 x=5

According to the second solution, we'll use the shortened multiplication formula:

(x−5)2=x2−10x+25=0 (x-5)^2=x^2-10x+25=0

We'll use the trinomial:

(x−5)(x−5)=0 (x-5)(x-5)=0

x−5=0 x-5=0

x=5 x=5

or

x−5=0 x-5=0

x=5 x=5

Therefore, according to all calculations, x=5 x=5

3

Final Answer

x=5 x=5

Key Points to Remember

Essential concepts to master this topic
  • Zero Property: If a2=0 a^2 = 0 , then a = 0
  • Technique: Take square root of both sides: (x−5)2=0 \sqrt{(x-5)^2} = \sqrt{0} gives x-5 = 0
  • Check: Substitute x = 5: (5−5)2=02=0 (5-5)^2 = 0^2 = 0 ✓

Common Mistakes

Avoid these frequent errors
  • Looking for two different solutions when equation equals zero
    Don't think (x−5)2=0 (x-5)^2 = 0 has two solutions like x = 2 and x = 5! A perfect square equals zero only when the base equals zero, giving one repeated solution. Always remember: if a2=0 a^2 = 0 , then a = 0 is the only solution.

Practice Quiz

Test your knowledge with interactive questions

Find the value of the parameter x.

\( (x-5)^2=0 \)

FAQ

Everything you need to know about this question

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

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When a perfect square equals zero, the expression inside the parentheses must equal zero. Since (x−5)2=0 (x-5)^2 = 0 , we need x-5 = 0, which gives us only one value: x = 5.

What if the equation was (x−5)2=4 (x-5)^2 = 4 instead?

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Then you'd have two solutions! Taking the square root: x-5 = ±2, so x = 7 or x = 3. But when the right side is zero, there's only one solution.

Can I expand (x−5)2 (x-5)^2 and solve as a quadratic?

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Yes, but it's unnecessary! Expanding gives x2−10x+25=0 x^2 - 10x + 25 = 0 , which factors back to (x−5)(x−5)=0 (x-5)(x-5) = 0 . Both factors give the same solution: x = 5.

How do I know when a quadratic has one repeated solution?

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Look for perfect square trinomials like x2−10x+25 x^2 - 10x + 25 , which factor as (x−5)2 (x-5)^2 . When both factors are identical, you get one repeated root.

What's the difference between one solution and two identical solutions?

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Mathematically, they're the same thing! We say x = 5 is a double root or repeated solution. The graph of y=(x−5)2 y = (x-5)^2 touches the x-axis at exactly one point.

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