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 to the inequality is \(-\frac{3}{2} \leq x \leq \frac{5}{7}\).

Step 1: Identify coefficients

For the quadratic inequality \(14x^2 + 11x - 15 \leq 0\), we need to first identify the coefficients \(a\), \(b\), and \(c\). In this equation, \(a = 14\), \(b = 11\), and \(c = -15\).

Step 2: Apply the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values for \(a\), \(b\), and \(c\) into the formula to find the roots. Here it becomes: \[ x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 14 \cdot (-15)}}{2 \cdot 14} \]

Step 3: Calculate the Discriminant

Calculate the discriminant, \(b^2 - 4ac\). Substitute the numbers: \[ 11^2 - 4 \cdot 14 \cdot (-15) = 121 + 840 = 961 \] This tells us the number of solutions the quadratic equation has.

Step 4: Solve for the Roots

With the discriminant calculated as 961, substitute it back into the formula: \[ x = \frac{-11 \pm \sqrt{961}}{28} \] Since \(\sqrt{961} = 31\), the roots become: \[ x_1 = \frac{-11 + 31}{28} = \frac{20}{28} = \frac{5}{7} \] \[ x_2 = \frac{-11 - 31}{28} = \frac{-42}{28} = -\frac{3}{2} \]

Step 5: Determine the Intervals for the Inequality

Using the roots \(-\frac{3}{2}\) and \(\frac{5}{7}\), determine tests in the intervals formed by these roots: 1. \(x < -\frac{3}{2}\)2. \(-\frac{3}{2} \leq x \leq \frac{5}{7}\)3. \(x > \frac{5}{7}\)

Step 6: Test the Intervals

Choose test points in each interval and substitute back into the original inequality: 1. For \(x = -2\) (\(x < -\frac{3}{2}\)), \(14(-2)^2 + 11(-2) - 15 = 56 - 22 - 15 = 19 > 0\)2. For \(x = 0\) (\(-\frac{3}{2} \leq x \leq \frac{5}{7}\)), \(14(0)^2 + 11(0) - 15 = -15 \leq 0\)3. For \(x = 1\) (\(x > \frac{5}{7}\)), \(14(1)^2 + 11(1) - 15 = 10 > 0\)

Step 7: Conclusion for the Solution

Since only the interval \(-\frac{3}{2} \leq x \leq \frac{5}{7}\) satisfies the inequality \(14x^2 + 11x - 15 \leq 0\), this is the solution to the inequality.

Key Concepts

Quadratic Formula

The quadratic formula is a foundational tool in solving quadratic equations and inequalities, like the one given in this exercise. It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use it, we first need to identify the coefficients of the quadratic equation. These coefficients, represented as \(a\), \(b\), and \(c\), are easily extracted from the standard form \(ax^2 + bx + c = 0\). In our exercise, they are \(a = 14\), \(b = 11\), and \(c = -15\). Substituting these into the formula helps find the values of \(x\) where the quadratic reaches zero, known as the roots.

Discriminant

Integral to using the quadratic formula is the discriminant, \(b^2 - 4ac\). This component, found within the square root of the quadratic formula, not only aids in solving for the roots but also reveals the nature of these roots.

• If the discriminant is positive, as in our case with \(961\), the quadratic equation has two distinct real roots.
• If it is zero, you get one real double root, indicating a perfect square trinomial.
• A negative discriminant means there are no real roots; instead, they are complex or imaginary.

For the given inequality, \(b^2 - 4 \times 14 \times (-15) = 961\), this confirms two distinct real roots exist.

Roots

Roots are the values of \(x\) that satisfy the quadratic equation when set to zero. They are essential in understanding the behavior of quadratic expressions, particularly when solving inequalities.After computing the discriminant and substituting it back into the quadratic formula, we solve for the roots:

• The first root, \(x_1 = \frac{-11 + 31}{28} = \frac{5}{7}\).
• The second root, \(x_2 = \frac{-11 - 31}{28} = -\frac{3}{2}\).

These roots are thresholds at which the quadratic expression changes its sign and they help demarcate the intervals to be tested in interval testing.

Interval Testing

To solve a quadratic inequality, interval testing helps determine where the inequality holds true. Here, the roots \(-\frac{3}{2}\) and \(\frac{5}{7}\) partition the number line into distinct intervals:

• \(x < -\frac{3}{2}\)
• \(-\frac{3}{2} \leq x \leq \frac{5}{7}\)
• \(x > \frac{5}{7}\)

Choosing a test point in each interval lets us verify whether the quadratic expression is less than or equal to zero in those regions.

• In the interval \(x < -\frac{3}{2}\), testing \(x = -2\) results in a positive value.
• For \(-\frac{3}{2} \leq x \leq \frac{5}{7}\), testing \(x = 0\) indicates the inequality holds.
• In \(x > \frac{5}{7}\), testing \(x = 1\) also results in a positive value.

From the testing, only the interval \(-\frac{3}{2} \leq x \leq \frac{5}{7}\) satisfies the inequality, giving us the solution to the problem.