Problem 2 - A Graphical Approach to College Algebra

Question

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 is \(x = 9\); \(x = 4\) is extraneous.

Step 1: Draw a Rough Sketch of the Functions

First, let's sketch the graphs of the two equations. The function \( y_1 = \sqrt{x} \) is a square root function that starts at the origin (0,0) and curves upwards to the right. Meanwhile, the linear function \( y_2 = x-6 \) is a straight line with a slope of 1, starting from \(-6\) on the y-axis. Draw these two graphs on the same coordinate plane to see where they might intersect.

Step 2: Set the Equations Equal to Each Other

Since we are looking for points where the two functions intersect, set \( y_1 = y_2 \). This gives us the equation \( \sqrt{x} = x - 6 \).

Step 3: Solve the Equation

To solve \( \sqrt{x} = x - 6 \), first isolate \( \sqrt{x} \) on one side if necessary (it's already isolated). Then, square both sides of the equation to eliminate the square root: \( x = (x - 6)^2 \).

Step 4: Expand and Rearrange the Equation

Expand \((x - 6)^2\) to get \(x = x^2 - 12x + 36\). Rearrange this into a standard form: \(0 = x^2 - 13x + 36\).

Step 5: Factor the Quadratic Equation

Factor the quadratic equation \(x^2 - 13x + 36 = 0\) into \((x - 4)(x - 9) = 0\).

Step 6: Solve for x

Set each factor equal to zero and solve for \(x\):- \(x - 4 = 0\) gives \(x = 4\)- \(x - 9 = 0\) gives \(x = 9\)

Step 7: Check for Extraneous Solutions

Check the solutions back in the original equations \(y_1 = \sqrt{x} \) and \( y_2 = x - 6 \) to verify them. For \(x = 4\):- \( \sqrt{4} = 2 \)- \( 4 - 6 = -2 \) (No match here, so \(x = 4\) is extraneous.)For \(x = 9\):- \( \sqrt{9} = 3 \)- \( 9 - 6 = 3 \) (These match, so \(x = 9\) is a valid solution.)

Step 8: State the Solution Set

The solution set for the equation where \(y_1 = y_2\) is \(x = 9\). The value \(x = 4\) is extraneous and does not satisfy both functions simultaneously.

Key Concepts

Quadratic Equations

Quadratic equations are polynomials of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The solutions to quadratic equations are the values of \( x \) that make the equation true. These solutions are often found by factoring, completing the square, or using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In this exercise, after setting \( \sqrt{x} = x - 6 \), squaring both sides results in a quadratic equation \( x^2 - 13x + 36 = 0 \). To solve this, factorization is used, resulting in the factors \( (x - 4)(x - 9) = 0 \). This leads to the potential solutions \( x = 4 \) and \( x = 9 \). Remember, finding a solution often involves solving a quadratic equation.

Graphing Functions

Graphing functions is a useful visual tool to understand the behavior and interaction of different equations. It involves plotting points that satisfy the equation on a coordinate plane.

The first function, \( y_1 = \sqrt{x} \), is a square root function. This function begins at the origin and curves upward to the right, representing a gradual increase. The second function, \( y_2 = x - 6 \), is a linear function with a slope of 1, beginning at \( y = -6 \).

By graphing these functions, you can see their shape and where they intersect, helping to visually confirm the solutions found algebraically. It helps when solving equations to have a mental picture of what's occurring on a graph.

Intersection of Functions

The intersection of functions occurs where the graphs of two equations meet or cross each other. This point represents the solution to the equation \( y_1 = y_2 \).

In this problem, \( \sqrt{x} = x - 6 \) signifies the intersection point of the graphs \( y_1 \) and \( y_2 \). By equating and solving \( \sqrt{x} = x - 6 \), we find the potential points of intersection. After solving, the intersections happen at \( x = 4 \) and \( x = 9 \) initially. However, only \( x = 9 \) is a valid solution upon verifying with the original equations.

Analyzing intersections helps confirm that the point truly lies on both graphs.

Extraneous Solutions

Extraneous solutions are results that emerge during the manipulation of equations but do not satisfy the original equation. It's important to double-check each potential solution to ensure it holds true for the initial problem.

In the given exercise, squaring the equation \( \sqrt{x} = x - 6 \) introduced the solution \( x = 4 \). However, upon verification, substituting \( x = 4 \) into the original equations shows inconsistencies: \( \sqrt{4} eq 4 - 6 \). Thus, \( x = 4 \) is considered an extraneous solution.

Always substitute the solutions back into the initial equations to differentiate valid solutions from extraneous ones.