Problem 72 - A Graphical Approach to College Algebra

Question

Solve each nonlinear system of equations analytically. $$\begin{aligned}&y=|x-1|\\\&y=x^{2}-4\end{aligned}$$

Step-by-Step Solution

Answer

Solutions: \( \left( \frac{1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} \right) \) and \( \left( \frac{-1 - \sqrt{21}}{2}, \frac{3 + \sqrt{21}}{2} \right) \).

Step 1: Understand Absolute Value Function

The first equation is \( y = |x-1| \). The absolute value function splits into two cases: \( y = x - 1 \) when \( x \geq 1 \) and \( y = -(x - 1) = 1 - x \) when \( x < 1 \).

Step 2: Case 1: Solve for \( x \geq 1 \)

For \( y = x - 1 \) where \( x \geq 1 \), set it equal to the second equation: \( x - 1 = x^2 - 4 \). Rearrange to form a quadratic equation: \( x^2 - x - 3 = 0 \).

Step 3: Solve Quadratic Equation for \( x \geq 1 \)

Use the quadratic formula to solve \( x^2 - x - 3 = 0 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -1, c = -3 \). Calculate \[ x = \frac{1 \pm \sqrt{13}}{2} \]. \( x \) must be \( \geq 1 \).

Step 4: Evaluate Solutions for \( x \geq 1 \)

\( x = \frac{1 + \sqrt{13}}{2} \) is approximately \( 2.302 \), which satisfies \( x \geq 1 \). Check \( y \): \( y = x - 1 = \frac{1 + \sqrt{13}}{2} - 1 = \frac{-1 + \sqrt{13}}{2} \). The point is \( \left( \frac{1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} \right) \).

Step 5: Case 2: Solve for \( x < 1 \)

For \( y = 1 - x \) where \( x < 1 \), set it equal to the second equation: \( 1 - x = x^2 - 4 \). Rearrange to form another quadratic: \( x^2 + x - 5 = 0 \).

Step 6: Solve Quadratic Equation for \( x < 1 \)

Use the quadratic formula for \( x^2 + x - 5 = 0 \): \[ x = \frac{-1 \pm \sqrt{1 + 20}}{2} \]. This simplifies to \( x = \frac{-1 \pm \sqrt{21}}{2} \). \( x \) must be \( < 1 \).

Step 7: Evaluate Solutions for \( x < 1 \)

\( x = \frac{-1 - \sqrt{21}}{2} \) is approximately \( -2.791 \), which satisfies \( x < 1 \). Check \( y \): \( y = 1 - x = 1 + \frac{1 + \sqrt{21}}{2} = \frac{3 + \sqrt{21}}{2} \). The solution is \( \left( \frac{-1 - \sqrt{21}}{2}, \frac{3 + \sqrt{21}}{2} \right) \).

Step 8: Combine Solutions

The solutions to the system are \( \left( \frac{1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} \right) \) and \( \left( \frac{-1 - \sqrt{21}}{2}, \frac{3 + \sqrt{21}}{2} \right) \).

Key Concepts

Absolute Value Function

An absolute value function is represented mathematically as \( y = |x| \). It measures the distance of a number from zero on the number line, disregarding direction. Consequently, it is always non-negative. When we operate on such a function with an expression like \( y = |x-1| \), we interpret it as the distance of \( x \) from 1. When dealing with absolute value functions in equations, we usually split the function into two separate linear equations to reflect the dual possibilities:

• One equation manages values greater than or equal to the reference point, \( x - 1 \), in our example.
• The other equation handles values less than the reference point, \( 1 - x \), accordingly.

This understanding is crucial for solving systems where one equation includes an absolute value function, as variations require evaluation on each interval independently.

Quadratic Equation

A quadratic equation is any equation equivalent to the form \( ax^2 + bx + c = 0 \). It characterizes a parabola and typically features an "\( x^2 \)" term, representing its highest degree. In the context of our example, rearranging terms after setting our linear expressions equal to \( x^2 - 4 \) results in two key quadratic equations:

• \( x^2 - x - 3 = 0 \)
• \( x^2 + x - 5 = 0 \)

Involving quadratic equations signals that a parabolic shape is part of the graph. Recognizing this form is essential since it offers insight into the solutions' nature. Special techniques, such as completing the square or utilizing the quadratic formula, are often used to disclose solutions due to quadratic complexity.

Quadratic Formula

To solve quadratic equations, the quadratic formula is a powerful tool. It's expressed as follows:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In our solution, the formula allows us to find \( x \)-values precisely by substituting values for \( a \), \( b \), and \( c \) from the quadratic equation. For example, with the equation \( x^2 - x - 3 = 0 \), it involves substituting:

• \( a = 1 \)
• \( b = -1 \)
• \( c = -3 \)

After calculation, we find \( x = \frac{1 \pm \sqrt{13}}{2} \). The "\( \pm \)" indicates potential two solutions, matching the case of quadratic equations naturally having up to two roots. Utilizing this approach can extract exact values when graphical methods or factorization isn't feasible. This dependable method usually forms part of advanced algebra discussions for its utility in navigating complex equations.