Problem 26 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ x^{2}+3 x+1>0 $$

Step-by-Step Solution

Answer

The solution set is \( x \in (-\infty, \frac{-3 - \sqrt{5}}{2}) \cup (\frac{-3 + \sqrt{5}}{2}, \infty) \).

Step 1: Identify the inequality

The given inequality is \[ x^2 + 3x + 1 > 0 \]

Step 2: Determine the roots of the quadratic equation

To solve the inequality, find the roots of the corresponding equation \[ x^2 + 3x + 1 = 0 \] using the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 3\), and \(c = 1\).

Step 3: Calculate the discriminant

Compute the discriminant of the equation: \[\Delta = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]

Step 4: Find the roots using the quadratic formula

Substitute the discriminant and the coefficients into the quadratic formula: \[ x = \frac{-3 \pm \sqrt{5}}{2} \] Hence, the roots are \[ x_1 = \frac{-3 + \sqrt{5}}{2} \] and \[ x_2 = \frac{-3 - \sqrt{5}}{2}. \]

Step 5: Test intervals determined by the roots

The roots \( \frac{-3 + \sqrt{5}}{2} \approx -0.382\) and \( \frac{-3 - \sqrt{5}}{2} \approx -2.618\) divide the number line into three intervals: 1. \( x < \frac{-3 - \sqrt{5}}{2} \) 2. \( \frac{-3 - \sqrt{5}}{2} < x < \frac{-3 + \sqrt{5}}{2} \) 3. \( x > \frac{-3 + \sqrt{5}}{2} \)

Step 6: Determine where the quadratic is positive

Choose a test point in each interval to determine the sign of \( x^2 + 3x + 1 \): 1. For \( x < \frac{-3 - \sqrt{5}}{2} \), choose \( x = -4 \): \[ (-4)^2 + 3(-4) + 1 = 16 - 12 + 1 = 5 > 0 \] 2. For \( \frac{-3 - \sqrt{5}}{2} < x < \frac{-3 + \sqrt{5}}{2} \), choose \( x = -1 \): \[ (-1)^2 + 3(-1) + 1 = 1 - 3 + 1 = -1 < 0 \] 3. For \( x > \frac{-3 + \sqrt{5}}{2} \), choose \( x = 1 \): \[ 1^2 + 3(1) + 1 = 1 + 3 + 1 = 5 > 0 \]

Step 7: Write the solution set

The solution to the inequality is the set of \(x\) values where \( x^2 + 3x + 1 > 0 \). This corresponds to the regions where the quadratic expression is positive: \[ x \in (-\infty, \frac{-3 - \sqrt{5}}{2}) \cup (\frac{-3 + \sqrt{5}}{2}, \infty) \]

Step 8: Illustrate the solution on the real number line

On the real number line, the solution is indicated by shading the regions \( (-\infty, \frac{-3 - \sqrt{5}}{2}) \) and \( (\frac{-3 + \sqrt{5}}{2}, \infty) \).

Key Concepts

quadratic equation

A quadratic equation is an equation of the form: \[ ax^2 + bx + c = 0 \] Here, 'a', 'b', and 'c' are constants with 'a' ≠ 0. This is because if 'a' were 0, then the equation would be linear, not quadratic. The solutions to the quadratic equation are the values of 'x' that make the equation true. These solutions, also known as roots, can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula works for any quadratic equation, whether it has two real roots, one real root, or complex roots. The term under the square root sign, \(b^2 - 4ac\), is called the discriminant, and it provides crucial information about the nature of the roots.

inequality solution

To solve a quadratic inequality, such as \( x^2 + 3x + 1 > 0 \), follow these steps:

• Identify the quadratic inequality.
• Find the roots of the corresponding quadratic equation by setting the inequality to equal zero.
• Use the quadratic formula to find these roots. These roots will divide the number line into intervals.
• Determine the sign of the quadratic expression in each interval by choosing test points within each interval.
• Identify the intervals where the inequality holds true.
• Write down the solution set and illustrate it on the real number line.

By following these steps, you ensure a rigorous and systematic approach to solving quadratic inequalities.

discriminant calculation

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the expression: \[ \Delta = b^2 - 4ac \]The discriminant helps in determining the nature of the roots of the quadratic equation:

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

To solve the given inequality \( x^2 + 3x + 1 > 0 \), the discriminant is calculated as follows:
\[ \Delta = 3^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]Since the discriminant is positive ( \( \Delta > 0 \) ), the quadratic equation has two distinct real roots. These roots then divide the number line into intervals, which are tested to find where the original inequality holds true.