Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the given quadratic inequality using the Quadratic Formula. $$ 14 x^{2}+11 x-15 \leq 0 $$

Step-by-Step Solution

Answer

The solution is \([-\frac{3}{2}, \frac{5}{7}]\).

Step 1: Identify the Quadratic Function Components

The given quadratic inequality is \( 14x^2 + 11x - 15 \leq 0 \). Identify the coefficients \( a = 14 \), \( b = 11 \), and \( c = -15 \). These will be used in the quadratic formula for finding the roots.

Step 2: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug in the coefficients into the formula: \( x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 14 \cdot (-15)}}{2 \cdot 14} \).

Step 3: Calculate the Discriminant

Compute the discriminant \( \Delta = b^2 - 4ac \). Here, \( \Delta = 11^2 - 4 \cdot 14 \cdot (-15) = 121 + 840 = 961 \). The discriminant is positive, meaning there are two real roots.

Step 4: Solve for the Roots

Substitute the discriminant back into the quadratic formula: \( x = \frac{-11 \pm \sqrt{961}}{28} \). Since \( \sqrt{961} = 31 \), we get the roots \( x_1 = \frac{-11 + 31}{28} = \frac{20}{28} = \frac{5}{7} \) and \( x_2 = \frac{-11 - 31}{28} = \frac{-42}{28} = -\frac{3}{2} \).

Step 5: Determine the Solution Interval

The roots \( x_1 = \frac{5}{7} \) and \( x_2 = -\frac{3}{2} \) divide the number line into intervals. Check the signs of the quadratic expression \( 14x^2 + 11x - 15 \) in the intervals \( ( -\infty, -\frac{3}{2} ) \), \( [-\frac{3}{2}, \frac{5}{7}] \), and \( (\frac{5}{7}, \infty) \) to find where it is non-positive.

Step 6: Test Each Interval

Choose a test point from each interval. For the interval \( (-\infty, -\frac{3}{2}) \), use \( x = -2 \). The quadratic is positive. For the interval \( [-\frac{3}{2}, \frac{5}{7}] \), use \( x = 0 \). The quadratic evaluates to \(-15\), which is non-positive. For \((\frac{5}{7}, \infty)\), use \( x = 1 \). The quadratic is positive. Therefore, the solution set is \([-\frac{3}{2}, \frac{5}{7}]\).

Key Concepts

Quadratic Formula

Understanding the quadratic formula is essential for solving quadratic equations and inequalities. The quadratic formula provides a solution for any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is written as:

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

This formula calculates the values of \( x \) that make the quadratic equation true. The formula uses the coefficients from the quadratic equation:

• \( a \) is the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

For the inequality \( 14x^2 + 11x - 15 \leq 0 \), we derive the values for \( a = 14 \), \( b = 11 \), and \( c = -15 \). Substituting these into the quadratic formula will guide us to find the roots of the equation. Calculating the roots using this formula is a structured way to handle quadratic problems.

Discriminant

The discriminant is a critical part of the quadratic formula that helps determine the nature of the roots. It is found inside the square root in the formula:

• \( \Delta = b^2 - 4ac \)

The value of the discriminant can tell us several things about the roots of the quadratic equation. Specifically, it informs us about the number and type of solutions:

• If \( \Delta > 0 \), there are two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root (also called a repeated or double root).
• If \( \Delta < 0 \), there are no real roots; the roots are complex.

In our problem, substituting \( a = 14 \), \( b = 11 \), and \( c = -15 \) into the discriminant formula gives \( \Delta = 961 \), which is greater than zero. Therefore, we have two real roots. Using the discriminant helps us understand that there are real solutions to the inequality, making it easier to determine intervals for the inequality solution.

Real Roots

Finding the real roots of the quadratic equation is crucial for solving quadratic inequalities. Since we calculated the discriminant \( \Delta = 961 \) to be positive, it indicates the presence of two real roots. Let's take this further by solving these roots with the quadratic formula:

• \( x_1 = \frac{-b + \sqrt{\Delta}}{2a} \)
• \( x_2 = \frac{-b - \sqrt{\Delta}}{2a} \)

By substituting our coefficients into these:

• For \( x_1 \), substituting the values yields \( \frac{-11 + 31}{28} = \frac{5}{7} \).
• For \( x_2 \), it is \( \frac{-11 - 31}{28} = -\frac{3}{2} \).

This means our real roots are \( x_1 = \frac{5}{7} \) and \( x_2 = -\frac{3}{2} \). These roots divide the number line into intervals critical for testing inequalities.

Inequality Solution

Solving a quadratic inequality involves determining the intervals where the inequality holds true. Given the quadratic inequality \( 14x^2 + 11x - 15 \leq 0 \), let's use the roots we found. These roots \( x_1 = \frac{5}{7} \) and \( x_2 = -\frac{3}{2} \) divide the number line.Our task is to check the sign of the quadratic expression in the intervals created by these roots:

• \( (-\infty, -\frac{3}{2}) \)
• \( [-\frac{3}{2}, \frac{5}{7}] \)
• \( (\frac{5}{7}, \infty) \)

To solve the inequality, we test points from each interval:

• For \( (-\infty, -\frac{3}{2}) \), test with \( x = -2 \), and observe the expression results positive.
• For \( [-\frac{3}{2}, \frac{5}{7}] \), test with \( x = 0 \). Here, \( 14(0)^2 + 11(0) - 15 = -15 \), which is non-positive.
• For \( (\frac{5}{7}, \infty) \), test with \( x = 1 \), resulting in a positive value.

As a result, the solution to the quadratic inequality is \( [-\frac{3}{2}, \frac{5}{7}] \), indicating where the quadratic holds true as non-positive across these intervals.