Solve the Quadratic Equation: (x-4)² - x(x+8) = 0

Q: Why did the x² terms cancel out in this problem?

Great observation! When we expanded $ (x-4)^2 $ we got $ x^2 - 8x + 16 $, and from $ -x(x+8) $ we got $ -x^2 - 8x $. The +x² and -x² cancelled each other out, leaving us with a linear equation!

Q: How do I remember the perfect square formula?

Think of it as "First squared, plus/minus twice the product, plus second squared": $ (a±b)^2 = a^2 ± 2ab + b^2 $. For $ (x-4)^2 $: first squared is $ x^2 $, twice the product is $ -2(x)(4) = -8x $, second squared is $ 16 $.

Q: What if I get a different answer when expanding?

Double-check each step! Common errors include: forgetting the negative sign in front of x(x+8), miscalculating $ 4^2 = 16 $, or not distributing the x to both terms inside the parentheses.

Q: Can I solve this without expanding first?

While you could try other methods, expansion is the most reliable approach here. It lets you see all terms clearly and combine like terms systematically, reducing the chance of errors.

Q: How do I know when I have a linear vs quadratic equation?

Look at the highest power of x after simplifying! If the highest power is 2, it's quadratic. If it's 1 (like in our final equation -16x = -16), it's linear. Sometimes quadratic-looking equations become linear after simplification!

Quadratic Equations with Expansion and Simplification

$(x-4)^2-x(x+8)=0$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:04 Use shortened multiplication formulas to expand brackets

00:17 Properly expand brackets, multiply by each factor

00:27 Collect like terms

00:35 Isolate X

00:49 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$(x-4)^2-x(x+8)=0$

Step-by-step solution

Let's solve the equation. First, we'll simplify the algebraic expressions using the perfect square binomial formula:

$(a\pm b)^2=a^2\pm2ab+b^2$ We'll apply the mentioned formula and expand the parentheses in the expressions in the equation:

$(x-4)^2-x(x+8)=0 \\ x^2-2\cdot x\cdot4+4^2-x^2-8x=0 \\ x^2-8x+16-x^2-8x=0$ In the first stage, we used the distributive property to expand the parentheses,

We'll continue and combine like terms, by moving terms between sides. Then - we can notice that the squared term cancels out and therefore it's a first-degree equation, which we'll solve by isolating the variable term on one side and dividing both sides of the equation by its coefficient:

$x^2-8x+16-x^2-8x=0 \\ -16x=-16\hspace{8pt}\text{/}:(-16)\\ \boxed{x=1}$ Therefore, the correct answer is answer D.

Final Answer

$x=1$

Key Points to Remember

Essential concepts to master this topic

Expansion: Apply distributive property and perfect square formula correctly
Technique: Combine like terms: -8x and -8x become -16x
Check: Substitute x=1: (1-4)² - 1(1+8) = 9 - 9 = 0 ✓

Common Mistakes

Avoid these frequent errors

Forgetting to expand both terms completely
Don't just expand (x-4)² and ignore distributing x(x+8) = wrong simplified equation! This leads to missing terms and incorrect solutions. Always expand every single term in the equation before combining like terms.

Practice Quiz

Test your knowledge with interactive questions

Declares the given expression as a sum

$$ (7b-3x)^2 $$

FAQ

Everything you need to know about this question

Why did the x² terms cancel out in this problem?

Great observation! When we expanded $(x-4)^2$ we got $x^2 - 8x + 16$ , and from $-x(x+8)$ we got $-x^2 - 8x$ . The +x² and -x² cancelled each other out, leaving us with a linear equation!

How do I remember the perfect square formula?

Think of it as "First squared, plus/minus twice the product, plus second squared": $(a±b)^2 = a^2 ± 2ab + b^2$ . For $(x-4)^2$ : first squared is $x^2$ , twice the product is $-2(x)(4) = -8x$ , second squared is $16$ .

What if I get a different answer when expanding?

Double-check each step! Common errors include: forgetting the negative sign in front of x(x+8), miscalculating $4^2 = 16$ , or not distributing the x to both terms inside the parentheses.

Can I solve this without expanding first?

While you could try other methods, expansion is the most reliable approach here. It lets you see all terms clearly and combine like terms systematically, reducing the chance of errors.

How do I know when I have a linear vs quadratic equation?

Look at the highest power of x after simplifying! If the highest power is 2, it's quadratic. If it's 1 (like in our final equation -16x = -16), it's linear. Sometimes quadratic-looking equations become linear after simplification!

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Short Multiplication Formulas questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime