Solve (x-4)² + x(x-12) = 16: Finding Parameter x

Quadratic Equations with Factoring Method

Find the value of the parameter x.

(x−4)2+x(x−12)=16 (x-4)^2+x(x-12)=16

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:10 Let's solve this problem together.
00:14 First, open the parentheses using multiplication formulas.
00:20 Make sure to multiply each term correctly.
00:25 Now, calculate the products and combine similar terms.
00:33 Next, simplify as much as possible.
00:44 Look for and factor out any common terms.
00:53 Set each factor equal to zero to find solutions.
00:58 You've found one solution! Use the same approach to find the se cond.
01:04 And that's the complete solution to the problem!

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−4)2+x(x−12)=16 (x-4)^2+x(x-12)=16

2

Step-by-step solution

Let's open the parentheses, remembering that there might be more than one solution for the value of X:

(x−4)2+x(x−12)=16 (x-4)^2+x(x-12)=16

x2−8x+16+x2−12x=16 x^2-8x+16+x^2-12x=16

2x2−20x=0 2x^2-20x=0

2x(x−10)=0 2x(x-10)=0

Therefore:

x−10=0 x-10=0

x=10 x=10

Or:

2x=0 2x=0

x=0 x=0

3

Final Answer

x=0,x=10 x=0,x=10

Key Points to Remember

Essential concepts to master this topic
  • Expansion: Use FOIL and distribute to get standard form
  • Factoring: Factor out common term: 2x(x−10)=0 2x(x-10)=0
  • Check: Substitute both solutions back: (0−4)2+0(0−12)=16 (0-4)^2 + 0(0-12) = 16 ✓

Common Mistakes

Avoid these frequent errors
  • Stopping after finding only one solution
    Don't solve just 2x=0 2x = 0 and forget x−10=0 x-10 = 0 = missing half the answer! Quadratic equations can have two solutions. Always set each factor equal to zero and solve both.

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 have two solutions?

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This is a quadratic equation (highest power is x2 x^2 ), and quadratic equations can have up to two solutions. When you factor to get 2x(x−10)=0 2x(x-10) = 0 , either factor can equal zero!

How do I expand (x−4)2 (x-4)^2 correctly?

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Remember that (x−4)2=(x−4)(x−4) (x-4)^2 = (x-4)(x-4) . Use FOIL: First + Outer + Inner + Last = x2−4x−4x+16=x2−8x+16 x^2 - 4x - 4x + 16 = x^2 - 8x + 16

What if I can't factor the quadratic?

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If factoring doesn't work easily, you can always use the quadratic formula. But in this case, factoring out the common factor 2x 2x makes it much simpler!

How do I check if both answers are correct?

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Substitute each solution back into the original equation. For x=0 x = 0 : (0−4)2+0(0−12)=16+0=16 (0-4)^2 + 0(0-12) = 16 + 0 = 16 ✓
For x=10 x = 10 : (10−4)2+10(10−12)=36+(−20)=16 (10-4)^2 + 10(10-12) = 36 + (-20) = 16 ✓

Why do we set the factored form equal to zero?

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This uses the Zero Product Property: if two factors multiply to give zero, then at least one factor must equal zero. So from 2x(x−10)=0 2x(x-10) = 0 , either 2x=0 2x = 0 or x−10=0 x-10 = 0 !

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