Problem 19 - College Algebra with Modeling and Visualization

Question

Solve each equation and inequality. Use set-builder or interval notation to write solution sets to the inequalities. (a) \(12 z^{2}-23 z+10=0\) (b) \(12 z^{2}-23 z+10 \leq 0\) (c) \(12 z^{2}-23 z+10 \geq 0\)

Step-by-Step Solution

Answer

(a) \(z = \frac{5}{4}, \frac{2}{3}\); (b) \([\frac{2}{3}, \frac{5}{4}]\); (c) \((-\infty, \frac{2}{3}] \cup [\frac{5}{4}, \infty)\)."

Step 1: Solve the Quadratic Equation

First, solve the quadratic equation \(12z^2 - 23z + 10 = 0\) using the quadratic formula, which is \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a = 12\), \(b = -23\), and \(c = 10\). The discriminant \(b^2 - 4ac\) is calculated as:\[(-23)^2 - 4 \times 12 \times 10 = 529 - 480 = 49\]The discriminant is a perfect square, so the equation has two real roots. Substitute into the quadratic formula:\[z = \frac{23 \pm \sqrt{49}}{24} = \frac{23 \pm 7}{24}\]So, the solutions are \(\frac{30}{24} = \frac{5}{4}\) and \(\frac{16}{24} = \frac{2}{3}\).

Step 2: Write the Solution to the Equation

The solutions to the equation \(12z^2 - 23z + 10 = 0\) are \(z = \frac{5}{4}\) and \(z = \frac{2}{3}\).

Step 3: Solve the Inequality using Roots

For the inequality \(12z^2 - 23z + 10 \leq 0\), we use the roots found previously, \(z = \frac{5}{4}\) and \(z = \frac{2}{3}\). These divide the number line into intervals to test:1. \((-\infty, \frac{2}{3})\)2. \((\frac{2}{3}, \frac{5}{4})\)3. \((\frac{5}{4}, \infty)\)We test a value from each interval to determine where the inequality holds:\- For \(z = 0\) in \((-\infty, \frac{2}{3})\), then \(12(0)^2 - 23(0) + 10 = 10 > 0\).- For \(z = 1\) in \((\frac{2}{3}, \frac{5}{4})\), then \(12(1)^2 - 23(1) + 10 = -1 \leq 0\).- For \(z = 2\) in \((\frac{5}{4}, \infty)\), then \(12(2)^2 - 23(2) + 10 = 18 > 0\).The inequality holds for the interval \(\left[\frac{2}{3}, \frac{5}{4}\right]\).

Step 4: Write the Solution in Interval Notation

The solution to the inequality \(12z^2 - 23z + 10 \leq 0\) in interval notation is \(\left[\frac{2}{3}, \frac{5}{4}\right]\).

Step 5: Solve the Second Inequality

For the inequality \(12z^2 - 23z + 10 \geq 0\), we want to find where the polynomial is greater than or equal to zero. Using the same intervals and tests as above:- \((-\infty, \frac{2}{3})\) where \(12(0)^2 - 23(0) + 10 = 10 > 0\).- \((\frac{2}{3}, \frac{5}{4})\) where \(12(1)^2 - 23(1) + 10 = -1 ot\geq 0\).- \((\frac{5}{4}, \infty)\) where \(12(2)^2 - 23(2) + 10 = 18 > 0\).The inequality holds for the intervals \(( -\infty, \frac{2}{3}] \cup [\frac{5}{4}, \infty)\).

Step 6: Write the Solution for the Second Inequality in Interval Notation

The solution to the inequality \(12z^2 - 23z + 10 \geq 0\) in interval notation is \((-\infty, \frac{2}{3}] \cup [\frac{5}{4}, \infty)\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation, which is any equation of the form \( ax^2 + bx + c = 0 \). To solve this, the quadratic formula is applied:

• \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this expression, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
The quadratic formula gives us two solutions: one with the plus sign (\(+\)) and one with the minus sign (\(-\)). These solutions correspond to the points where the parabola represented by the quadratic equation crosses the x-axis.

In the exercise, we applied the quadratic formula to \( 12z^2 - 23z + 10 = 0 \). Here, \( a = 12 \), \( b = -23 \), and \( c = 10 \). After calculating the discriminant \( b^2 - 4ac \) and finding it was a perfect square, we could easily find the roots:

• \( z = \frac{5}{4} \)
• \( z = \frac{2}{3} \)

This tells us the values of \( z \) that make the original equation equal zero.

Inequalities

When solving inequalities, such as \( 12z^2 - 23z + 10 \leq 0 \), you're looking for the range of values for \( z \) that make the inequality true. Unlike equations, inequalities are about ranges or intervals of numbers rather than specific solutions.
To solve it, we first identified the roots of the related equation \( 12z^2 - 23z + 10 = 0 \), which are \( \frac{5}{4} \) and \( \frac{2}{3} \). These roots help divide the number line into different intervals that need to be tested individually:

• \((-\infty, \frac{2}{3})\)
• \((\frac{2}{3}, \frac{5}{4})\)
• \((\frac{5}{4}, \infty)\)

To determine which intervals satisfy the inequality, we substituted test points in each interval back into the polynomial:

• \( z = 0 \) (Interval: \((-\infty, \frac{2}{3})\)) results in a positive value, not satisfying \( \leq 0 \).
• \( z = 1 \) (Interval: \((\frac{2}{3}, \frac{5}{4})\)) results in a negative value, satisfying \( \leq 0 \).
• \( z = 2 \) (Interval: \((\frac{5}{4}, \infty)\)) results in a positive value, not satisfying \( \leq 0 \).

Finally, we found that the inequality holds true in the interval \([\frac{2}{3}, \frac{5}{4}]\), where \( 12z^2 - 23z + 10 \leq 0 \).

Interval Notation

Interval notation is a compact way to represent the set of solutions to an inequality. It's particularly useful for expressing ranges of numbers concisely.
In interval notation, we use brackets and parentheses to convey whether endpoints of the interval are included or not:

• "[" or "]" denotes that an endpoint is included.
• "(" or ")" indicates that an endpoint is not included.

For instance, the inequality solution for \( 12z^2 - 23z + 10 \leq 0 \) was written in interval notation as \([\frac{2}{3}, \frac{5}{4}]\). Here, both endpoints are enclosed by square brackets, meaning both \( \frac{2}{3} \) and \( \frac{5}{4} \) are included in the solution set.

Meanwhile, for the inequality \( 12z^2 - 23z + 10 \geq 0 \), the solution in interval notation was \((-\infty, \frac{2}{3}] \cup [\frac{5}{4}, \infty)\). This shows two separate intervals combined (union) – the first ends at \( \frac{2}{3} \) and the second begins at \( \frac{5}{4} \). The infinity symbols and round brackets indicate that the intervals extend in the respective indefinite direction without including the infinite points themselves.