Solve (x-4)² = (x+2)(x-1): Comparing Squared and Factored Forms

Q: Why does this quadratic equation become linear after expanding?

Great observation! When we expand both sides, the $ x^2 $ terms cancel out: $ x^2 - 8x + 16 = x^2 + x - 2 $. This leaves us with only linear terms, making it a first-degree equation.

Q: Do I always need to expand both sides first?

Yes! Always expand both sides completely before moving terms. This ensures you don't miss any terms and can properly combine like terms. Trying to solve without expanding leads to errors.

Q: How do I remember the perfect square formula?

Think of it as First² - 2(First)(Last) + Last². For $ (x-4)^2 $: $ x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16 $. Practice with simple numbers first!

Q: What if I get confused during the expansion?

Take it step by step! Write out each expansion separately:$ (x-4)^2 = x^2 - 8x + 16 $$ (x+2)(x-1) = x^2 + x - 2 $Then set them equal and combine.

Q: How can I check if x = 2 is really correct?

Substitute x = 2 into the original equation:Left side: $ (2-4)^2 = (-2)^2 = 4 $Right side: $ (2+2)(2-1) = (4)(1) = 4 $Since 4 = 4, our answer is correct!

Quadratic Equations with Binomial Expansions

$(x-4)^2=(x+2)(x-1)$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:03 We'll use shortened multiplication formulas to open the parentheses

00:09 Open parentheses properly, multiply each factor by each factor

00:21 Calculate the multiplications

00:27 Collect like terms

00:32 Simplify what we can

00:44 Isolate X

01:11 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

$(x-4)^2=(x+2)(x-1)$

Step-by-step solution

Let's examine the given equation:

$(x-4)^2=(x+2)(x-1)$ First, let's simplify the equation, using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$ and the expanded distributive law,

We'll start by opening the parentheses using the perfect square binomial formula mentioned and using the expanded distributive law and then we'll move terms and combine like terms:

$(x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-2\cdot x\cdot4+4^2=x^2-x+2x-2\\ x^2-8x+16=x^2-x+2x-2\\ x^2-8x+16-x^2+x-2x+2=0\\ -9x+18=0$ We got a first-degree equation, we'll solve it in the regular way by isolating the variable on one side:

$-9x+18=0 \\ -9x=-18\hspace{6pt}\text{/}:(-9)\\ \boxed{x=2}$

Let's summarize the equation solving steps:

$(x-4)^2=(x+2)(x-1) \\ \downarrow\\ x^2-8x+16=x^2-x+2x-2\\ -9x+18=0 \\ \downarrow\\ \boxed{x=2}$ Therefore, the correct answer is answer C.

Final Answer

$x=2$

Key Points to Remember

Essential concepts to master this topic

Expansion Rule: Apply $(a-b)^2 = a^2 - 2ab + b^2$ and distributive property
Technique: Expand both sides: $x^2 - 8x + 16 = x^2 + x - 2$
Check: Substitute x = 2: $(2-4)^2 = 4$ and $(2+2)(2-1) = 4$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to expand one side completely
Don't expand just $(x-4)^2$ and leave $(x+2)(x-1)$ as is = incomplete equation! This prevents proper term cancellation and combining. Always expand both sides completely before moving terms to one side.

Practice Quiz

Test your knowledge with interactive questions

$$ x^2-3x-18=0 $$

FAQ

Everything you need to know about this question

Why does this quadratic equation become linear after expanding?

Great observation! When we expand both sides, the $x^2$ terms cancel out: $x^2 - 8x + 16 = x^2 + x - 2$ . This leaves us with only linear terms, making it a first-degree equation.

Do I always need to expand both sides first?

Yes! Always expand both sides completely before moving terms. This ensures you don't miss any terms and can properly combine like terms. Trying to solve without expanding leads to errors.

How do I remember the perfect square formula?

Think of it as First² - 2(First)(Last) + Last². For $(x-4)^2$ : $x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16$ . Practice with simple numbers first!

What if I get confused during the expansion?

Take it step by step! Write out each expansion separately:

$(x-4)^2 = x^2 - 8x + 16$
$(x+2)(x-1) = x^2 + x - 2$

Then set them equal and combine.

How can I check if x = 2 is really correct?

Substitute x = 2 into the original equation:
Left side: $(2-4)^2 = (-2)^2 = 4$
Right side: $(2+2)(2-1) = (4)(1) = 4$
Since 4 = 4, our answer is correct!

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