Mirror Framing: Find Frame Thickness With Quadratics

Hey guys! Let's dive into a cool math problem that combines geometry and algebra. Imagine Alexei, who's got this awesome boat and wants to hang a mirror in it. He also wants to add a frame around the mirror, making it look super stylish. Now, here's the puzzle: the total area of the mirror and frame needs to be 19.25 square feet. The mirror itself is 3 feet wide and 5 feet long. Our mission? To figure out which quadratic equation we can use to find the thickness of the frame.

Setting up the Problem: Mirror, Frame, and Area

Alright, let's break this down step by step. We've got a rectangular mirror that's 3 feet by 5 feet. When we add a frame around it, we're essentially increasing the dimensions of the mirror. Think of the frame as an extra border of uniform thickness, which we'll call 'x', around the mirror. This means the frame extends 'x' feet on all sides of the mirror.

To really nail this, it's crucial to visualize what's happening. Picture the mirror as the inner rectangle, and the frame as the area surrounding it. The frame adds to both the width and the length of the mirror. If the frame is 'x' feet thick, it adds 'x' feet on each side, so we're adding 2x to both the width and the length. This is a key concept in understanding how the dimensions change and how we'll set up our equation.

So, what are the new dimensions of the entire setup (mirror + frame)? The original width of the mirror is 3 feet, and we're adding 'x' feet on each side, making the new width (3 + 2x) feet. Similarly, the original length of the mirror is 5 feet, and adding 'x' feet on each side gives us a new length of (5 + 2x) feet. Remember, the frame adds to the dimensions in both directions, hence the '2x'.

Now, we know the total area of the mirror and the frame combined is 19.25 square feet. Area, for a rectangle, is simply length times width. So, the area of the entire setup can be expressed as (3 + 2x)(5 + 2x). And we know this area equals 19.25 square feet. This gives us the foundation for our equation, and we're well on our way to finding that quadratic equation!

Building the Quadratic Equation

Now, let's translate our understanding into a mathematical equation. We've established that the total area of the mirror and frame is (3 + 2x)(5 + 2x) square feet, and we know this area is equal to 19.25 square feet. So, we can write our equation as:

(3 + 2x)(5 + 2x) = 19.25

This equation is the heart of our problem. It represents the relationship between the frame's thickness ('x') and the total area. But it's not in the standard quadratic form yet. To get there, we need to expand the left side of the equation.

Expanding the product (3 + 2x)(5 + 2x) involves using the distributive property (often remembered by the acronym FOIL - First, Outer, Inner, Last). Let's go through it step by step:

• First: Multiply the first terms in each parenthesis: 3 * 5 = 15
• Outer: Multiply the outer terms: 3 * 2x = 6x
• Inner: Multiply the inner terms: 2x * 5 = 10x
• Last: Multiply the last terms: 2x * 2x = 4x²

Adding these together, we get: 15 + 6x + 10x + 4x². Now, let's combine like terms (the 'x' terms) to simplify: 6x + 10x = 16x. So, our expanded equation looks like this:

4x² + 16x + 15 = 19.25

We're getting closer to the standard quadratic form, which is ax² + bx + c = 0. To achieve this, we need to move the 19.25 from the right side of the equation to the left side. We do this by subtracting 19.25 from both sides:

4x² + 16x + 15 - 19.25 = 0

Now, let's simplify the constant terms: 15 - 19.25 = -4.25. This gives us our final quadratic equation:

4x² + 16x - 4.25 = 0

This is the quadratic equation we can use to determine the thickness of the frame. It's in the standard form, and we can now use various methods (like the quadratic formula, factoring, or completing the square) to solve for 'x', the thickness of the frame.

Solving for the Frame's Thickness: Methods and Considerations

Now that we've got our quadratic equation, 4x² + 16x - 4.25 = 0, the next step is to solve for 'x', which represents the thickness of the frame. There are several methods we can use to solve quadratic equations, each with its own strengths and best-use cases. Let's explore a couple of the most common ones and discuss which might be most suitable for this problem.

1. The Quadratic Formula: A Reliable Workhorse

The quadratic formula is a foolproof method that works for any quadratic equation, regardless of whether it can be easily factored. It's a bit like a Swiss Army knife – always reliable. The formula is:

x = (-b ± √(b² - 4ac)) / (2a)

Where 'a', 'b', and 'c' are the coefficients from our quadratic equation in the standard form (ax² + bx + c = 0). In our case, a = 4, b = 16, and c = -4.25. Let's plug these values into the formula:

x = (-16 ± √(16² - 4 * 4 * -4.25)) / (2 * 4)

Now, we simplify step by step:

x = (-16 ± √(256 + 68)) / 8 x = (-16 ± √324) / 8 x = (-16 ± 18) / 8

This gives us two possible solutions:

x₁ = (-16 + 18) / 8 = 2 / 8 = 0.25 x₂ = (-16 - 18) / 8 = -34 / 8 = -4.25

2. Factoring: When Simplicity Shines

Factoring is another method, but it's most effective when the quadratic equation can be factored easily. It involves rewriting the quadratic expression as a product of two binomials. However, in our case, with the coefficients 4, 16, and -4.25, factoring might not be the most straightforward approach. The numbers aren't as "clean" as we'd like for easy factoring, and it might involve some trial and error.

3. Completing the Square: A Method for Deeper Understanding

Completing the square is a method that involves manipulating the equation to form a perfect square trinomial. It's a bit more involved than the quadratic formula but provides a deeper understanding of the structure of quadratic equations. While it's a valuable technique, it might be more time-consuming for this particular problem compared to the quadratic formula.

Choosing the Right Solution and Practical Considerations

We've found two potential solutions for 'x' using the quadratic formula: 0.25 and -4.25. But here's where we need to apply some common sense. Remember, 'x' represents the thickness of the frame. Can the thickness be a negative value? Nope! A negative thickness doesn't make sense in the real world.

Therefore, we discard the negative solution, -4.25, and our valid solution is x = 0.25. This means the frame is 0.25 feet thick. That's 3 inches (since 0.25 feet * 12 inches/foot = 3 inches). So, Alexei needs a frame that's 3 inches thick to make the total area of the mirror and frame 19.25 square feet.

Real-World Application and Why This Matters

This problem might seem like a purely mathematical exercise, but it actually has practical applications in various real-world scenarios. Think about it: framing a picture, building a border around a garden, or even designing the layout of a room. All these situations involve calculating areas and dimensions, and sometimes, you need to solve a quadratic equation to get the desired result.

In Alexei's case, he needed to figure out the frame thickness to achieve a specific total area. This is a common problem in woodworking, construction, and design. Understanding how to set up and solve quadratic equations can help you make accurate calculations and avoid costly mistakes in these fields.

Moreover, this problem highlights the importance of connecting math concepts to real-life situations. Math isn't just about abstract numbers and formulas; it's a powerful tool that can help us solve practical problems. When we can see the relevance of math in our daily lives, it becomes more engaging and meaningful.

So, next time you're faced with a similar problem, remember the steps we took: visualize the situation, set up the equation, solve for the unknown, and interpret the solution in the context of the problem. You'll be surprised at how useful math can be!

Key Takeaways and Final Thoughts

Let's wrap up what we've learned from Alexei's mirror-framing adventure. This problem was a great example of how quadratic equations can be used to solve real-world problems involving areas and dimensions. We started with a geometric scenario, translated it into an algebraic equation, and then used the quadratic formula to find the solution.

Here are some key takeaways from this problem:

• Visualizing the problem is crucial. Drawing a diagram or picture can help you understand the relationships between the different quantities.
• Setting up the equation correctly is essential. Make sure you understand how the frame affects the dimensions of the mirror and how the area is calculated.
• The quadratic formula is a powerful tool. It can be used to solve any quadratic equation, even if it's not easily factorable.
• Always interpret the solutions in context. Make sure your answer makes sense in the real world. In this case, we had to discard the negative solution because a frame thickness cannot be negative.
• Math is relevant to real-life situations. This problem showed how quadratic equations can be used in woodworking, construction, design, and other fields.

By working through this problem, we've not only sharpened our math skills but also gained a better appreciation for how math connects to the world around us. So, keep practicing, keep visualizing, and keep applying math to solve problems – you'll be amazed at what you can achieve!