Problem 2 - A Graphical Approach to Precalculus with Limits

Question

In Exercises begin by drawing a rough sketch to determine the number of real solutions for the equation \(y_{1}=y_{2}\) Then solve this equation by hand. Give the solution set and any extraneous values that may occur. Do not use a calculator. $$\begin{aligned} &y_{1}=\sqrt{x}\\\ &y_{2}=x-6 \end{aligned}$$

Step-by-Step Solution

Answer

The solution set is \(\{9\}\); \(x = 4\) is extraneous.

Step 1: Sketch the Functions

To determine the number of real solutions for the equation \(y_1 = y_2\), we need to draw a rough sketch of both functions. \(y_1 = \sqrt{x}\) represents the square root function, which starts at the origin (0,0) and increases slowly to the right. \(y_2 = x - 6\) is a linear function with a slope of 1, intersecting the y-axis at -6, and increases linearly.

Step 2: Find Intersection Points

We set \(\sqrt{x} = x - 6\) to find intersections, meaning the points where the graphs of the functions intersect. We need to solve this equation for the values of \(x\) that make both sides equal.

Step 3: Solve the Equation Algebraically

To solve \(\sqrt{x} = x - 6\), first remove the square root by squaring both sides: \((\sqrt{x})^2 = (x - 6)^2\). This gives us \(x = x^2 - 12x + 36\). Rearranging terms gives \(x^2 - 13x + 36 = 0\).

Step 4: Use the Quadratic Formula

The quadratic equation \(x^2 - 13x + 36 = 0\) can be solved using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -13\), \(c = 36\). Calculate the discriminant: \((-13)^2 - 4 \times 1 \times 36 = 169 - 144 = 25\). Hence, \(x = \frac{13 \pm 5}{2}\), giving solutions \(x = 9\) and \(x = 4\).

Step 5: Verify Solutions and Check for Extraneous Values

Verify these solutions in the original equations. For \(x = 9\), \(y_1 = \sqrt{9} = 3\) and \(y_2 = 9 - 6 = 3\), which match, hence \(x = 9\) is a valid solution. For \(x = 4\), \(y_1 = \sqrt{4} = 2\) and \(y_2 = 4 - 6 = -2\), which do not match. Therefore, \(x = 4\) is extraneous.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides a direct way to find the solutions (or roots) of the equation. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula works because it is derived from completing the square, a method that transforms any quadratic equation into a perfect square trinomial, which can then be easily solved. Here’s why each component is important:

• a, b, and c: These are the coefficients from the quadratic equation.
• \(b^2 - 4ac\): Known as the discriminant, determines the nature of the roots. If it's positive, there are two distinct real solutions. If zero, there is one real solution, and if negative, the solutions are complex.
• \(-b \pm \): The plus-minus sign indicates there are usually two solutions.

In our example, we used the quadratic formula on the equation \(x^2 - 13x + 36 = 0\), finding \(x = 9\) and \(x = 4\), but verifying later alerted us to an extraneous solution at \(x = 4\). This step highlights the importance of not just solving equations but also verifying solutions in the context of the original problem.

Graphical Solution

A graphical solution involves plotting equations to visually determine their intersections, which correspond to the solutions of the equation \(y_1 = y_2\). In the original problem, sketching is a valuable first step.For \(y_1 = \sqrt{x}\) and \(y_2 = x - 6\):

• \(y_1 = \sqrt{x}\): Starts at (0,0) and rises slowly to the right, defining a curve.
• \(y_2 = x - 6\): A straight line with a slope of 1 and a y-intercept at -6, moving upwards.

Sketching helps us determine potential solution points by visualizing where the graphs intersect. In this case, the intersection suggests possible solutions at \(x = 9\) and \(x = 4\). Once these intersections are identified, it's crucial to validate them by substituting back into the original equations, confirming their accuracy. Graphical methods give a good intuitive idea but should always be paired with algebraic solutions to ensure thorough correctness.

Extraneous Solutions

Extraneous solutions are potential answers to an equation that arise during the solution process but are not valid when substituted back into the context of the original problem. They often appear when manipulations involve powers or roots, which can inadvertently introduce "false" solutions.In our problem, solving \(\sqrt{x} = x - 6\) algebraically by squaring both sides led to the quadratic equation \(x^2 - 13x + 36 = 0\). Solving this indeed provided two potential solutions: \(x = 9\) and \(x = 4\). Here’s why they were checked:

• \(x = 9\): Substitution into the original equation confirmed it as a valid solution. Both sides equaled 3.
• \(x = 4\): When substituted back, it resulted in both sides being unequal (2 and -2), showing this to be an extraneous solution.

Extraneous solutions remind us that while algebraic processes yield solutions, the context of the problem affects their validity. Always check each solution back in the original equation to confirm its correctness.