Chapter 1: Problem 13
Express the solution set of the given inequality in interval notation and sketch its graph. $$ 2 x^{2}+5 x-3>0 $$
Short Answer
Expert verified
The solution set is \((-\infty, -3) \cup (\frac{1}{2}, \infty)\).
Step by step solution
01
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 \(c = -3\).
02
Calculate the discriminant
First, calculate the discriminant \(\Delta = b^2 - 4ac\). Substitute the values: \(\Delta = 5^2 - 4 \times 2 \times (-3) = 25 + 24 = 49\).
03
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 \(x_2 = \frac{-12}{4} = -3\).
04
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.
05
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 \((\frac{1}{2}, \infty)\): \(2(1)^2 + 5(1) - 3 = 2 + 5 - 3 = 4 > 0\).
06
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.
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
- \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
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.
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 \(\frac{1}{2}\).
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.