Problem 106 - A Graphical Approach to College Algebra

Question

Solve each inequality analytically. Support your answers graphically. Give exact values for endpoints. $$(a) 3 x^{2}+13 x+10 \leq 0$$ $$(b) 3 x^{2}+13 x+10>0$$

Step-by-Step Solution

Answer

(a) \(-\frac{10}{3} \leq x \leq -1\); (b) \(x < -\frac{10}{3}\) or \(x > -1\).

Step 1: Solve the Quadratic Equation

We first solve the quadratic equation associated with both inequalities. The equation to solve is:\[3x^2 + 13x + 10 = 0\]We use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 3\), \(b = 13\), and \(c = 10\). Compute the discriminant:\[b^2 - 4ac = 13^2 - 4 \cdot 3 \cdot 10 = 169 - 120 = 49\]Now, apply the quadratic formula:\[x = \frac{-13 \pm \sqrt{49}}{6} = \frac{-13 \pm 7}{6}\]This gives:\[x = \frac{-6}{6} = -1\] and \[x = \frac{-20}{6} = -\frac{10}{3}\].These are the solutions of the quadratic equation, which will serve as endpoints.

Step 2: Analyze the Inequalities

Now that we have the roots \(-1\) and \(-\frac{10}{3}\), determine the sign of the expression on intervals defined by these roots:For inequality (a) \(3x^2 + 13x + 10 \leq 0\), we test intervals:\- \((-\infty, -\frac{10}{3})\)- \((-\frac{10}{3}, -1)\)- \((-1, \infty)\) Choosing test points, such as \(-4\), \(-2\), and \(0\), evaluate the quadratic in these intervals to find which are true for \(\leq 0\).For inequality (b) \(3x^2 + 13x + 10 > 0\), determine which intervals satisfy the condition through similar testing as above.

Step 3: Solving Inequality (a)

Test the intervals:- For \((-\infty, -\frac{10}{3})\), choose \(x = -4\): \[3(-4)^2 + 13(-4) + 10 = 48 - 52 + 10 = 6 > 0\] (False)- For \((-\frac{10}{3}, -1)\), choose \(x = -2\): \[3(-2)^2 + 13(-2) + 10 = 12 - 26 + 10 = -4 \leq 0\] (True)- For \((-1, \infty)\), choose \(x = 0\): \[3(0)^2 + 13(0) + 10 = 10 > 0\] (False)Thus, the solution for (a) is \(-\frac{10}{3} \leq x \leq -1\).

Step 4: Solving Inequality (b)

Evaluate the intervals again:- For \((-\infty, -\frac{10}{3})\), as tested before: \(x = -4\) gives positive value (True).- For \((-\frac{10}{3}, -1)\), as tested before: \(x = -2\) gives negative value (False).- For \((-1, \infty)\), as tested before: \(x = 0\) gives positive value (True).Thus, the solution for (b) is \(x < -\frac{10}{3}\) or \(x > -1\).

Step 5: Graphical Representation

To support the analytical analysis, plot the quadratic function \(y = 3x^2 + 13x + 10\). The parabola opens upwards (since \(a = 3 > 0\)). The roots \(-\frac{10}{3}\) and \(-1\) represent the x-intercepts.For inequality (a), shade the region between the roots \([-\frac{10}{3}, -1]\), inclusive.For inequality (b), shade outside these roots: \((\infty, -\frac{10}{3})\) and \((-1, \infty)\).

Key Concepts

Solving Quadratic Equations

To solve a quadratic inequality, we first need to find the roots by solving the related quadratic equation. Here, we consider the equation: \[3x^2 + 13x + 10 = 0\] To find the solutions, we use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where \(a = 3\), \(b = 13\), and \(c = 10\). First, calculate the discriminant: \[b^2 - 4ac = 13^2 - 4 \cdot 3 \cdot 10 = 49\] Since the discriminant is positive, the equation has two real roots. Plug in the values: \[x = \frac{-13 \pm \sqrt{49}}{6} = \frac{-13 \pm 7}{6}\] This results in two solutions:

• \(x = -1\)
• \(x = -\frac{10}{3}\)

These roots are important because they help us define the intervals for testing the inequalities.

Graphical Representation

Understanding the nature of the quadratic functions graphically can vastly simplify determining which intervals satisfy an inequality. The graph of a quadratic function, \[y = 3x^2 + 13x + 10\] is a parabola. Because the coefficient \(a\) is positive, the parabola opens upward.The roots of the quadratic function \(-\frac{10}{3}\) and \(-1\) are the x-intercepts where the graph crosses the x-axis. These intercepts split the x-axis into three intervals:

• \((-\infty, -\frac{10}{3})\)
• \((-\frac{10}{3}, -1)\)
• \((-1, \infty)\)

By placing these points on a graph, we observe that the parabola is below the x-axis between \(-\frac{10}{3}\) and \(-1\), indicating where the quadratic expression holds less than or equal to zero. To understand inequality solutions thoroughly, it helps to visualize and confirm them graphically.

Inequality Intervals

With the quadratic solving and graphical understanding, we now explore inequality intervals. For the quadratic inequality \(3x^2 + 13x + 10 \leq 0\), we test the intervals created by the roots:

• Interval \((-\infty, -\frac{10}{3})\): Test a value like \(x = -4\). The inequality becomes false here because the value is positive.
• Interval \((-\frac{10}{3}, -1)\): Choose \(x = -2\). The result is negative, satisfying \(\leq 0\).
• Interval \((-1, \infty)\): Test \(x = 0\). The outcome is positive, hence false for \(\leq 0\).

For \(3x^2 + 13x + 10 > 0\), the inequalities are true in the outside regions:

• Interval \((-\infty, -\frac{10}{3})\): Tested with \(x = -4\), yielding a positive result.
• Interval \((-\frac{10}{3}, -1)\): With \(x = -2\), results are negative, hence false.
• Interval \((-1, \infty)\): Using \(x = 0\), condition is true because the outcome is positive.

Each interval test confirms where the quadratic expression satisfies the specified inequality, thus providing a full understanding of the solution set.