Problem 1 - 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. $$\begin{aligned} &y_{1}=\sqrt{x}\\\ &y_{2}=2 x-1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 1\), with no extraneous values.

Step 1: Understand the Functions

The first function given is \(y_1 = \sqrt{x}\), which is the square root function. The second function is \(y_2 = 2x - 1\), which is a linear equation. We need to find where these two functions are equal, i.e., where their graphs intersect.

Step 2: Sketch the Graphs

Draw a rough sketch of the graphs of \(y_1 = \sqrt{x}\) and \(y_2 = 2x - 1\). 1. The graph of \(y_1 = \sqrt{x}\) starts at the origin \((0,0)\) and rises in a curve moving rightward.2. The graph of \(y_2 = 2x - 1\) is a straight line with a slope of 2. It intersects the y-axis at -1.Look for points where the graphs might potentially intersect.

Step 3: Set the Functions Equal

To find the intersection points, set the equations equal to each other: \[\sqrt{x} = 2x - 1\] Solve this equation to find possible solutions for \(x\).

Step 4: Solve the Equation

Square both sides of the equation to eliminate the square root:\[(\sqrt{x})^2 = (2x - 1)^2\] This simplifies to:\[x = 4x^2 - 4x + 1\]Rearrange to form a quadratic equation:\[4x^2 - 5x + 1 = 0\]

Step 5: Use the Quadratic Formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to the equation \(4x^2 - 5x + 1 = 0\), where \(a = 4\), \(b = -5\), and \(c = 1\). Calculate:1. \(b^2 - 4ac = (-5)^2 - 4(4)(1) = 25 - 16 = 9\).2. \(x = \frac{-(-5) \pm \sqrt{9}}{2(4)} = \frac{5 \pm 3}{8}\).This gives the solutions:\[x = \frac{5 + 3}{8} = 1\]\[x = \frac{5 - 3}{8} = \frac{1}{4}\]

Step 6: Check for Extraneous Solutions

Check these solutions in the original equations to confirm they are valid:1. For \(x = 1\), \(y_1 = \sqrt{1} = 1\) and \(y_2 = 2(1) - 1 = 1\). Both match.2. For \(x = \frac{1}{4}\), \(y_1 = \sqrt{\frac{1}{4}} = \frac{1}{2}\) and \(y_2 = 2\left(\frac{1}{4}\right) - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\). Values do not match here, hence it's extraneous.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in algebra for solving quadratic equations. These equations take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The solutions for \( x \) in a quadratic equation can be determined using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the roots of the equation by finding where the parabola intersects the x-axis. Understanding each part of the formula:

• \( -b \): The coefficient of \( x \) negated.
• \( \sqrt{b^2 - 4ac} \): Called the discriminant, it determines the number and type of solutions.
• \( 2a \): Duplicates the leading coefficient to form the denominator.

The discriminant \( b^2 - 4ac \) plays a crucial role:

• If \( > 0 \), there are two distinct real solutions.
• If \( = 0 \), there is one double real solution.
• If \( < 0 \), the solutions are complex and involve imaginary numbers.

Using this formula, students can solve any quadratic equation, like the one in the original problem, and predict the nature of its solutions.

Square Root Function

The square root function, represented by \( y = \sqrt{x} \), is a fundamental function in mathematics. This type of function only deals with non-negative inputs since the square root of a negative number results in an imaginary number.The graph of the square root function:- Begins at the point \((0, 0)\).- As \(x\) increases, the function values also increase.- Forms a curve that rises slowly and steadily to the right.Some important attributes of the square root function:

• Domain: All non-negative numbers (\(x \geq 0\)).
• Range: All non-negative numbers (\(y \geq 0\)).
• Intercept: The function intersects the y-axis at the origin.

The square root function is crucial for solving equations where a square root is present, like \(y_1 = \sqrt{x}\) in our problem. Understanding its graph helps predict where it might intersect with other functions, such as linear equations.

Linear Equation

Linear equations, such as \( y = mx + b \), represent the simplest form of equations where the relationship between \(x\) and \(y\) is a straight line. In the formula, \( m \) is the slope and \( b \) is the y-intercept.For the linear equation \( y_2 = 2x - 1 \):- Slope \( m = 2 \) implies that for every unit increase in \( x \), \(y\) increases by 2 units.- Y-intercept \(-1\) indicates that the line crosses the y-axis at \( (0, -1) \).Linear equations have a constant rate of change, resulting in a graph that's a straight line:

• The slope \( m \) determines the steepness and direction of the line.
• The y-intercept \( b \) provides a starting point on the y-axis.

In our exercise, understanding the graph of a linear equation like \( y_2 = 2x - 1 \) helps us set it equal to the square root function. By identifying where these lines intersect, students can solve for solutions where the two equations are equal. This is a powerful technique for determining solution sets in algebra.