Chapter 6: Problem 72
Solve each nonlinear system of equations analytically. $$\begin{aligned}&y=|x-1|\\\&y=x^{2}-4\end{aligned}$$
Short Answer
Expert verified
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 by step solution
01
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 \( x < 1 \).
02
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: \( x^2 - x - 3 = 0 \).
03
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 \( \geq 1 \).
04
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 \( \left( \frac{1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} \right) \).
05
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: \( x^2 + x - 5 = 0 \).
06
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 \( < 1 \).
07
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 \( \left( \frac{-1 - \sqrt{21}}{2}, \frac{3 + \sqrt{21}}{2} \right) \).
08
Combine Solutions
The solutions to the system are \( \left( \frac{1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} \right) \) and \( \left( \frac{-1 - \sqrt{21}}{2}, \frac{3 + \sqrt{21}}{2} \right) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
- \( x^2 - x - 3 = 0 \)
- \( x^2 + x - 5 = 0 \)
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:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -3 \)