Problem 52 - Algebra for College Students

Question

Set up an equation and solve each problem. A 5-inch by 7-inch picture is surrounded by a frame of uniform width. The area of the picture and frame together is 80 square inches. Find the width of the frame.

Step-by-Step Solution

Answer

The width of the frame is 1.5 inches.

Step 1: Define the Variables

Let's assign a variable to the unknown quantity. We need to find the width of the frame, so let's call this width "x".

Step 2: Set Up the Equation

The dimensions of the picture are 5 inches by 7 inches. If the frame has a uniform width of "x", the total width becomes \(5 + 2x\) inches and the total height becomes \(7 + 2x\) inches. Hence, the equation for the area (picture + frame) is: \ \[(5 + 2x)(7 + 2x) = 80\]

Step 3: Expand the Equation

Expand the left side of the equation to simplify it: \ \[(5 + 2x)(7 + 2x) = 35 + 10x + 14x + 4x^2 = 80\] \ Combine like terms: \ \[4x^2 + 24x + 35 = 80\]

Step 4: Move All Terms to One Side

Subtract 80 from both sides to set the equation to zero: \ \[4x^2 + 24x + 35 - 80 = 0\] \ This simplifies to: \ \[4x^2 + 24x - 45 = 0\]

Step 5: Solve the Quadratic Equation

To find the values of x, we use the quadratic formula: \ \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] \ where \(a = 4\), \(b = 24\), and \(c = -45\). Calculate the discriminant: \ \[b^2 - 4ac = 24^2 - 4(4)(-45) = 576 + 720 = 1296\] \ Now, solve using the quadratic formula: \ \[x = \frac{-24 \pm \sqrt{1296}}{8}\] \ \[x = \frac{-24 \pm 36}{8}\] \ This gives two solutions: \ \[x = \frac{-24 + 36}{8} = 1.5\] \ and \ \[x = \frac{-24 - 36}{8} = -7.5\].

Step 6: Choose the Valid Solution

Since a width cannot be negative, we discard \(-7.5\) and keep \(x = 1.5\) inches as the valid solution.

Key Concepts

Area Calculation

To understand how to find the width of the frame, we first need to focus on the idea of area calculation. In geometry, area refers to the measure of space within a closed boundary or shape. For rectangular shapes, such as pictures or frames, the area is calculated by multiplying its length by its width.

In this exercise, we are given a picture measuring 5 inches by 7 inches. When we add a frame that surrounds the picture, we need to consider the extra width from the frame on each side. Therefore, the width and height of the entire framed picture are both increased by twice the width of the frame ("x").

Thus, the total dimensions become:

• Width: \(5 + 2x\) inches
• Height: \(7 + 2x\) inches

The complete area (inside picture plus the frame) is then the product of these dimensions, which is set equal to 80 square inches. This forms our fundamental equation for solving this problem.

Algebraic Expressions

Algebraic expressions allow us to represent mathematical concepts using symbols and variables. In this problem, the unknown variable is the width of the frame, which we have labeled as "x". By using algebraic expressions, we can create a formula that represents the area of the framed picture.

The expression \((5 + 2x)(7 + 2x)\) symbolizes the area of the entire framed picture. Expanding this expression is key to simplifying the problem and consists of applying the distributive property (also known as the FOIL method in binomials):

• First: \(5 \times 7 = 35\)
• Outer: \(5 \times 2x = 10x\)
• Inner: \(2x \times 7 = 14x\)
• Last: \(2x \times 2x = 4x^2\)

After combining like terms from this expression, we end up with \(4x^2 + 24x + 35\). This is an important step in our problem-solving process as it converts a complex figure into a solvable quadratic equation.

Problem-Solving in Algebra

Solving the quadratic equation derived from our area display is the heart of problem-solving in algebra. Quadratic equations generally take the form \(ax^2 + bx + c = 0\). In this scenario, our equation \(4x^2 + 24x - 45 = 0\) fits this standard format.

To find "x", we employ the quadratic formula, given by:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

"a", "b", and "c" are coefficients representing numbers in our equation. The key steps involved in utilizing this formula include:

• Calculating the discriminant, \(b^2 - 4ac\)
• Substituting these values into the formula to solve for "x"

This solution process yields two potential results for "x". Since width cannot be a negative number, only the positive solution (1.5 inches) is valid. This logical deduction is crucial in confirming our solution to the problem.