Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Express the solution set of the given inequality in interval notation and sketch its graph. $$ 2 x^{2}+5 x-3>0 $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -3) \cup (\frac{1}{2}, \infty)\).

Step 1: Find the roots of the equation

To solve the inequality, start by finding the roots of the quadratic equation associated with it: \(2x^2 + 5x - 3 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = 5\), and \(c = -3\).

Step 2: Calculate the discriminant

First, calculate the discriminant \(\Delta = b^2 - 4ac\). Substitute the values: \(\Delta = 5^2 - 4 \times 2 \times (-3) = 25 + 24 = 49\).

Step 3: Solve for the roots

Since the discriminant is 49, which is a perfect square, the roots are real and rational. Substitute \(\Delta = 49\) into the quadratic formula: \(x = \frac{-5 \pm \sqrt{49}}{4}\). This simplifies to \(x = \frac{-5 \pm 7}{4}\). The roots are \(x_1 = \frac{2}{4} = \frac{1}{2}\) and \(x_2 = \frac{-12}{4} = -3\).

Step 4: Analyze the sign of the quadratic expression

The parabola described by \(y = 2x^2 + 5x - 3\) opens upwards, and its roots divide the number line into intervals. Consider the intervals: \((-\infty, -3)\), \((-3, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). Test a point from each interval to determine where the inequality \(2x^2 + 5x - 3 > 0\) holds true.

Step 5: Test intervals

Test \(x = -4\) in \((-\infty, -3)\): \(2(-4)^2 + 5(-4) - 3 = 32 - 20 - 3 = 9 > 0\).Test \(x = 0\) in \((-3, \frac{1}{2})\): \(2(0)^2 + 5(0) - 3 = -3 < 0\).Test \(x = 1\) in \((\frac{1}{2}, \infty)\): \(2(1)^2 + 5(1) - 3 = 2 + 5 - 3 = 4 > 0\).

Step 6: Write the solution in interval notation

The inequality \(2x^2 + 5x - 3 > 0\) holds for the intervals where the tested values give positive results. This is for \(x \in (-\infty, -3) \cup (\frac{1}{2}, \infty)\). The solution set is the union of these intervals.

Step 7: Sketch the graph

Draw a number line. Mark the critical points \(-3\) and \(\frac{1}{2}\). Shade the intervals \((-\infty, -3)\) and \((\frac{1}{2}, \infty)\) to indicate where the inequality holds.

Key Concepts

Quadratic Formula

To solve quadratic inequalities like \(2x^2 + 5x - 3 > 0\), we first need to find the roots of the quadratic equation, \(2x^2 + 5x - 3 = 0\). For this, the quadratic formula is a handy tool. The formula is given by:

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

Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation \(ax^2 + bx + c = 0\). Substituting the values of \(a = 2\), \(b = 5\), and \(c = -3\) into the formula allows us to calculate the roots. These roots indicate where the parabola intersects the x-axis. Knowing these intersections helps in determining the intervals where the inequality holds true. This formula comes in particularly helpful for equations that cannot be factored easily, providing an exact answer for any quadratic. After computing with the formula, you'll find specific points at which the parabola hits the x-axis, thus allowing further analysis of the inequality.

Discriminant

The discriminant, represented by \(\Delta = b^2 - 4ac\), provides crucial details about the nature of the roots of a quadratic equation. For the given equation, the discriminant value (\(\Delta = 49\)) is calculated by substituting \(b = 5\), \(a = 2\), and \(c = -3\).

• If \(\Delta > 0\), the equation has two distinct real roots.
• If \(\Delta = 0\), there is exactly one real root, as the parabola touches the x-axis at this point.
• If \(\Delta < 0\), the roots are complex with no real intersection on the x-axis.

Since the discriminant here is 49 and it's a perfect square, it confirms that there are two distinct, real, and rational roots for the equation. By determining the discriminant first, you can quickly assess how the roots of the quadratic equation behave which directly influences the solution of inequalities.

Interval Notation

When expressing solutions of inequalities, interval notation provides a simplified and structured format to indicate ranges of numbers. In the context of the inequality \(2x^2 + 5x - 3 > 0\), after identifying the roots \(-3\) and \(\frac{1}{2}\), these points divide the number line into distinct intervals.

• \((-fty, -3)\) signifies numbers less than \(-3\).
• \((-3, \frac{1}{2})\) represents numbers between \(-3\) and \(\frac{1}{2}\).
• \((\frac{1}{2}, \infty)\) includes all numbers greater than \(\frac{1}{2}\).

For this inequality, testing the sign of each interval helps determine where the quadratic expression is greater than zero. Solutions expressible in interval notation, such as \((-\infty, -3) \cup (\frac{1}{2}, \infty)\), clearly shows the union of intervals where the inequality holds true. This notation style efficiently communicates the solution, especially in exams or mathematical discussions.

Graphing Inequalities

Understanding quadratic inequalities benefits significantly from graphical representation. Graphing the quadratic function \(y = 2x^2 + 5x - 3\) aids in visualizing where the function resides above the x-axis. For inequalities such as \(2x^2 + 5x - 3 > 0\), these are areas where the graph is above the x-axis.

• The graph is a parabola, and since the leading term is positive, it opens upwards.
• The roots, \(-3\) and \(\frac{1}{2}\), indicate where the graph touches the x-axis.

By plotting these roots and the intervals they form, shading occurs in the areas where the inequality is true. Doing so helps students see the solutions visually, creating a better understanding of which intervals the parabola is positive. Graphing complements algebraic methods, fostering a deeper grasp of quadratic inequalities.