Problem 67 - Precalculus Student Solutions Manual 5th

Question

For \(x^{2}-x\) to have an absolute value equal to \(6,\) what are the two possible values that it may be? (Hint: One is positive and the other is negative.)

Step-by-Step Solution

Answer

The possible values of \(x\) are 3 and -2.

Step 1: Understand the Absolute Value Equation

The given expression is \(x^{2} - x\). We need to find the values of \(x\) such that the absolute value \(|x^{2} - x|\) is equal to 6. This translates to two separate equations: \(x^{2} - x = 6\) and \(x^{2} - x = -6\).

Step 2: Solve for \(x\) when \(x^{2} - x = 6\)

Rewrite \(x^{2} - x = 6\) as a standard quadratic equation: \(x^{2} - x - 6 = 0\). Factorize the equation to \((x-3)(x+2) = 0\). This gives the solutions \(x = 3\) and \(x = -2\).

Step 3: Solve for \(x\) when \(x^{2} - x = -6\)

Rewrite \(x^{2} - x = -6\) as a standard quadratic equation: \(x^{2} - x + 6 = 0\). To solve this quadratic equation, use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Substitute \(a = 1, b = -1, c = 6\): \[x = \frac{1 \pm \sqrt{1 - 24}}{2} = \frac{1 \pm \sqrt{-23}}{2}\]. Since \sqrt{-23}\ is not a real number, this quadratic equation has no real solutions.

Step 4: Confirm the Real Solutions

The values of \(x\) that make \(x^{2} - x = 6\) are \(x = 3\) and \(x = -2\). Since there are no real solutions for \(x^{2} - x = -6\), the final solutions are \(x = 3\) and \(x = -2\).

Key Concepts

quadratic equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). It forms a parabolic curve when graphed. Quadratic equations are fundamental in algebra because they appear in various real-life applications, from physics to engineering.

Quadratic equations can be solved using different methods, such as completing the square, factoring, or applying the quadratic formula. The solutions to quadratic equations are the points where the parabola intersects the x-axis, known as the roots of the equation.

factoring

Factoring is a method used to solve quadratic equations by expressing the quadratic expression as a product of its factors. For instance, the equation \( x^2 - x - 6 = 0 \) can be factored into \( (x - 3)(x + 2) = 0 \).

Here's a brief breakdown of the steps:

• Identify two numbers that multiply to give the constant term (in this case, -6), and add to give the middle coefficient (in this case, -1).
• Those numbers are 3 and -2, as \( 3 \times -2 = -6 \) and \( 3 - 2 = 1 \).
• Rewrite the quadratic equation using these numbers: \( (x - 3)(x + 2) = 0 \).

Factoring simplifies the quadratic equation and allows us to find the solutions by setting each factor equal to zero: \( x - 3 = 0 \) gives \( x = 3 \), and \( x + 2 = 0 \) gives \( x = -2 \). These solutions show where the parabola crosses the x-axis.

quadratic formula

The quadratic formula is a powerful tool for solving any quadratic equation: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. It is derived from the process of completing the square and applies to any quadratic equation of the form \( ax^2 + bx + c = 0 \).

To use the formula, simply identify the values of \( a \), \( b \), and \( c \) from the quadratic equation. Substitute them into the formula, and you'll be able to find the roots of the equation. For example, using the quadratic formula for \( x^2 - x + 6 = 0 \), we substitute \( a = 1, b = -1, \) and \( c = 6 \): \[ x = \frac{1 \pm \sqrt{1 - 24}}{2} = \frac{1 \pm \sqrt{-23}}{2} \]. Since the discriminant (\( b^2 - 4ac \)) is negative, it shows there are no real solutions, only complex ones.

real and complex solutions

Quadratic equations can have real or complex solutions, based on the value of the discriminant (\( b^2 - 4ac \)) in the quadratic formula.

• If the discriminant is positive, the equation has two distinct real solutions.
• If the discriminant is zero, there is one real solution (a repeated root).
• If the discriminant is negative, there are no real solutions, only complex solutions.

In our exercise, the discriminant of \( x^2 - x - 6 = 0 \) is positive, so it has two real solutions: \( x = 3 \) and \( x = -2 \). For \( x^2 - x + 6 = 0 \), the discriminant is negative, leading to no real solutions. Instead, it has complex solutions given by \( \frac{1 \pm \sqrt{-23}}{2} \). Understanding whether solutions are real or complex helps us interpret the nature of the roots and the graph of the quadratic equation.