Calculate X in Rectangle Area: Simplifying (3x+4)(3x-4)

Q: How do I recognize when to use difference of squares?

Look for the pattern (something + number)(same something - same number). When you see this, like (3x+4)(3x-4), you can skip FOIL and go straight to a² - b²!

Q: Why does x have two answers (±3)?

When we solve $ x^2 = 9 $, both positive and negative values work: 3² = 9 and (-3)² = 9. Always include the ± symbol for square root solutions!

Q: Can the area of a rectangle be negative?

No, areas are always positive! But the variable x itself can be negative. When we substitute x = -3, we get $ 9(-3)^2 - 16 = 81 - 16 = 65 $, which is positive.

Q: Do I always get perfect square solutions?

Not always! This problem gave us $ x^2 = 9 $, which is a perfect square. Sometimes you might get $ x^2 = 7 $ and need to write $ x = ±\sqrt{7} $.

Difference of Squares with Rectangle Applications

The area of the rectangle below is equal to: $(3x+4)(3x-4)$ .

Calculate a x.

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:09 Let's find X!

00:13 Use the formula for the rectangle's area. It's side times side.

00:19 Now, plug in the values you have to find the area.

00:23 That's the area of the rectangle.

00:27 Next, use this area in your equation. Solve for X.

00:39 Let's expand the brackets using multiplication rules.

00:57 Open the brackets and calculate three squared and four squared.

01:10 Now, isolate X in the equation.

01:26 Take the square root to get possible answers for X.

01:32 And there you have it! That's how to solve the problem.

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

The area of the rectangle below is equal to: $(3x+4)(3x-4)$ .

Calculate a x.

Step-by-step solution

To solve this problem, we start by recognizing that the expression for the area of the rectangle is given by the formula $(3x+4)(3x-4)$ . This can be simplified using the difference of squares:

$(3x+4)(3x-4) = (3x)^2 - (4)^2 = 9x^2 - 16$ .

The problem also provides the dimensions of the rectangle as 5 and 13. The area of the rectangle can therefore also be calculated as $5 \times 13 = 65$ .

We set the two expressions for the area equal to each other to find $x$ :

$9x^2 - 16 = 65$ .

Next, we solve for $x$ :

$9x^2 - 16 = 65$
$9x^2 = 65 + 16$
$9x^2 = 81$
$x^2 = \frac{81}{9}$
$x^2 = 9$
$x = \pm 3$ .

Therefore, the value of $x$ is $\operatorname{\pm}3$ .

Final Answer

$\operatorname{\pm}3$

Key Points to Remember

Essential concepts to master this topic

Pattern: (a+b)(a-b) = a² - b² for difference of squares
Technique: (3x+4)(3x-4) = (3x)² - 4² = 9x² - 16
Check: Substitute x = ±3: (3·3+4)(3·3-4) = 13×5 = 65 ✓

Common Mistakes

Avoid these frequent errors

Expanding (3x+4)(3x-4) using FOIL instead of difference of squares
Don't use FOIL: (3x+4)(3x-4) = 9x² - 12x + 12x - 16 = wrong steps! This wastes time with unnecessary middle terms that cancel anyway. Always recognize the difference of squares pattern (a+b)(a-b) = a² - b² immediately.

Practice Quiz

Test your knowledge with interactive questions

Solve:

$$ (2+x)(2-x)=0 $$

FAQ

Everything you need to know about this question

How do I recognize when to use difference of squares?

Look for the pattern (something + number)(same something - same number). When you see this, like (3x+4)(3x-4), you can skip FOIL and go straight to a² - b²!

Why does x have two answers (±3)?

When we solve $x^2 = 9$ , both positive and negative values work: 3² = 9 and (-3)² = 9. Always include the ± symbol for square root solutions!

Can the area of a rectangle be negative?

No, areas are always positive! But the variable x itself can be negative. When we substitute x = -3, we get $9(-3)^2 - 16 = 81 - 16 = 65$ , which is positive.

What if the rectangle dimensions were different?

The method stays the same! Set your expanded expression equal to length × width. For example, if dimensions were 7 and 10, you'd solve $9x^2 - 16 = 70$ instead.

Do I always get perfect square solutions?

Not always! This problem gave us $x^2 = 9$ , which is a perfect square. Sometimes you might get $x^2 = 7$ and need to write $x = ±\sqrt{7}$ .

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