Problem 49 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation. For equations with real solutions, support your answers graphically. $$(3-x)^{2}=25$$

Step-by-Step Solution

Answer

The solutions are \(x = -2\) and \(x = 8\).

Step 1: Isolate the Square Term

To solve the equation \((3-x)^{2}=25\), first recognize that the expression on the left is a square term. We start by isolating this square term: the equation is already in a suitable form.

Step 2: Take the Square Root

Next, take the square root of both sides of the equation to eliminate the square: \(\sqrt{(3-x)^{2}} = \sqrt{25}\), which simplifies to \(|3-x| = 5\).

Step 3: Solve the Absolute Value Equation

The equation \(|3-x| = 5\) means that \(3-x = 5\) or \(3-x = -5\). These represent two separate linear equations.

Step 4: Solve the First Linear Equation

First, solve \(3-x = 5\). Subtract 3 from both sides to get \(-x = 2\). Then multiply by -1 to isolate \(x\): \(x = -2\).

Step 5: Solve the Second Linear Equation

Next, solve \(3-x = -5\). Subtract 3 from both sides to get \(-x = -8\). Then multiply by -1 to isolate \(x\): \(x = 8\).

Step 6: Verify Solutions Graphically

We can verify these solutions by graphing the left and right sides of the original equation separately: \((3-x)^2\) and 25. The solutions \(x = -2\) and \(x = 8\) correspond to the x-values where these graphs intersect.

Key Concepts

Graphical Representation

To understand the graphical representation of the equation \((3-x)^{2}=25\), we can visualize the problem as finding where two curves intersect on a graph. Graphically, this involves plotting two functions:

• \(y = (3-x)^2\)
• \(y = 25\)

These are a downward-opening parabola and a horizontal line, respectively. The curve \(y = (3-x)^2\) shifts the standard parabola \(y = x^2\) down and to the right by 3 units. The line \(y = 25\) is constant and horizontal. To find solutions, we look for the x-values at which the parabola and the line intersect.
At these intersecting points, the values of the two functions are equal, solving the equation. For our problem, the solutions occur where these two graphs meet, which can be visualized as the x-coordinates \(-2\) and \(8\). Each indicate a point at which the parabola reaches the height of 25.

Square Root Method

The square root method is a straightforward and efficient way to solve equations involving squared terms. In the given equation \((3-x)^{2}=25\), the first step is to isolate the squared term, which is already done for us. The next step involves taking the square root of both sides of the equation. This yields:

• \(\sqrt{(3-x)^{2}} = \sqrt{25}\)
• \(|3-x| = 5\)

Taking the square root introduces an absolute value, because both positive and negative roots can equal the original squared number. It's crucial to consider both solutions because squaring either value results in the same original number.
Consequently, we obtain two scenarios to solve:

• \(3-x = 5\)
• \(3-x = -5\)

These lead us to two linear equations, necessary for finding all potential solutions of the equation.

Linear Equations

After using the square root method, we're left with two simpler linear equations to solve. Linear equations have the general form \(Ax + B = 0\), with a constant rate of change. Our derived equations:

• \(3-x = 5\)
• \(3-x = -5\)

require solving for \(x\). Starting with \(3-x=5\):

• Subtract 3 from both sides to get \(-x=2\).
• Multiply by -1 to isolate \(x\), resulting in \(x=-2\).

For \(3-x=-5\):

• Subtract 3 from both sides to achieve \(-x=-8\).
• Multiply by -1 to solve for \(x\), giving \(x=8\).

Each solution of a linear equation represents a specific point where the equation holds true. These x-values, \(x=-2\) and \(x=8\), are our solutions, determined from the intersections on the graph, verifying the original quadratic equation.