Solve The Following Quadratic Inequalities: 1) X² - 4x + 6 > 0, 2) X² + 6x + 10 < 0, 3) X² + X + 2 > 0, 4) X² + 3x + 5 < 0, 5) 2x² - 3x + 7 < 0, 6) 4x² - 8x + 9 > 0, And 7) -3.6x² - 7.2x < 0.
Quadratic inequalities, a fascinating area of mathematics, extend the concept of quadratic equations by introducing inequality signs. These inequalities, expressed in the general form of ax² + bx + c > 0 (or <, ≥, ≤), where a, b, and c are constants and a ≠ 0, play a crucial role in various mathematical and real-world applications. Understanding how to solve these inequalities is essential for students and professionals alike. This article delves into solving several quadratic inequalities, providing step-by-step explanations and insights into the underlying principles. We will explore techniques for determining the solution sets, which represent the range of x-values that satisfy the given inequality. This exploration involves analyzing the discriminant, finding roots, and interpreting the parabolic nature of quadratic functions to effectively solve each inequality. Mastering these techniques not only enhances mathematical proficiency but also provides a foundation for tackling more complex problems in calculus, optimization, and other advanced fields. By the end of this article, you will have a comprehensive understanding of how to approach and solve quadratic inequalities, equipping you with the skills to confidently handle a variety of mathematical challenges.
1) Solving x² - 4x + 6 > 0
To solve the quadratic inequality x² - 4x + 6 > 0, we first analyze the quadratic expression. The inequality asks for the values of x for which the quadratic expression is positive. One effective approach is to analyze the discriminant and identify the vertex of the parabola represented by the quadratic. The discriminant, denoted as Δ, is given by the formula Δ = b² - 4ac, where a, b, and c are the coefficients of the quadratic equation ax² + bx + c = 0. In this case, a = 1, b = -4, and c = 6. Plugging these values into the formula, we get:
Δ = (-4)² - 4(1)(6) = 16 - 24 = -8.
Since the discriminant is negative, Δ < 0, the quadratic equation x² - 4x + 6 = 0 has no real roots. This implies that the parabola represented by the quadratic function does not intersect the x-axis. To determine whether the quadratic expression is always positive or always negative, we consider the coefficient a, which is the coefficient of x². Here, a = 1, which is positive. This indicates that the parabola opens upwards. Given that the parabola opens upwards and does not intersect the x-axis (due to the negative discriminant), the quadratic expression is always positive for all real values of x. Therefore, the solution to the inequality x² - 4x + 6 > 0 is all real numbers. In interval notation, this is represented as (-∞, ∞).
In summary, to solve this inequality, we followed these steps: first, we computed the discriminant to determine the nature of the roots; second, we observed the sign of the coefficient a to understand the orientation of the parabola; and third, we deduced the solution based on these observations. This systematic approach is crucial for solving various quadratic inequalities effectively.
2) Solving x² + 6x + 10 < 0
Next, we address the quadratic inequality x² + 6x + 10 < 0. This inequality challenges us to find values of x for which the quadratic expression is negative. As before, we will utilize the discriminant and vertex analysis to determine the solution. The coefficients of the quadratic equation x² + 6x + 10 = 0 are a = 1, b = 6, and c = 10. We calculate the discriminant Δ using the formula Δ = b² - 4ac:
Δ = (6)² - 4(1)(10) = 36 - 40 = -4.
With a negative discriminant, Δ < 0, the quadratic equation x² + 6x + 10 = 0 has no real roots. This signifies that the parabola represented by the quadratic function does not intersect the x-axis. Since the coefficient a = 1 is positive, the parabola opens upwards. Consequently, the quadratic expression is always positive, as the parabola is entirely above the x-axis. Thus, there are no real values of x that make the expression x² + 6x + 10 negative. The solution set for the inequality x² + 6x + 10 < 0 is therefore empty, which we can denote as the empty set, ∅.
In essence, the absence of real roots and the upward-opening nature of the parabola dictate that the quadratic expression is never less than zero. This process reinforces the understanding that analyzing the discriminant and the sign of a are critical steps in solving quadratic inequalities. The empty set solution highlights the importance of recognizing cases where no real values satisfy the inequality.
3) Solving x² + x + 2 > 0
For the inequality x² + x + 2 > 0, we once again examine the discriminant to understand the nature of the roots. Here, a = 1, b = 1, and c = 2. The discriminant Δ is calculated as follows:
Δ = (1)² - 4(1)(2) = 1 - 8 = -7.
As the discriminant is negative (Δ < 0), the quadratic equation x² + x + 2 = 0 has no real roots. This indicates that the parabola does not intersect the x-axis. The coefficient a = 1 is positive, which means the parabola opens upwards. Since the parabola opens upwards and has no real roots, the quadratic expression is always positive for any real value of x. Therefore, the solution to the inequality x² + x + 2 > 0 is the set of all real numbers, represented in interval notation as (-∞, ∞). This outcome reinforces the principle that a positive a and a negative discriminant imply that the quadratic expression is always positive.
4) Solving x² + 3x + 5 < 0
Let's consider the inequality x² + 3x + 5 < 0. To solve this, we calculate the discriminant. With a = 1, b = 3, and c = 5, the discriminant Δ is given by:
Δ = (3)² - 4(1)(5) = 9 - 20 = -11.
The discriminant is negative (Δ < 0), indicating that the quadratic equation x² + 3x + 5 = 0 has no real roots. The coefficient a = 1 is positive, meaning the parabola opens upwards. Since the parabola does not intersect the x-axis and opens upwards, the quadratic expression is always positive. Therefore, there are no real values of x that make x² + 3x + 5 less than 0. The solution set is the empty set, ∅. This instance demonstrates that an upward-opening parabola with no real roots results in a quadratic expression that is always positive, thereby having no solution for the < 0 inequality.
5) Solving 2x² - 3x + 7 < 0
Now, let’s analyze the quadratic inequality 2x² - 3x + 7 < 0. In this case, a = 2, b = -3, and c = 7. The discriminant Δ is calculated as:
Δ = (-3)² - 4(2)(7) = 9 - 56 = -47.
The discriminant is negative (Δ < 0), which means the quadratic equation 2x² - 3x + 7 = 0 has no real roots. The coefficient a = 2 is positive, so the parabola opens upwards. As the parabola does not intersect the x-axis and opens upwards, the quadratic expression is always positive. Thus, there are no real values of x for which 2x² - 3x + 7 is less than 0. The solution set is the empty set, denoted as ∅. This situation further exemplifies how a positive leading coefficient and a negative discriminant lead to a quadratic expression that is always positive, resulting in no solution for inequalities requiring negative values.
6) Solving 4x² - 8x + 9 > 0
For the inequality 4x² - 8x + 9 > 0, we compute the discriminant with a = 4, b = -8, and c = 9:
Δ = (-8)² - 4(4)(9) = 64 - 144 = -80.
The discriminant is negative (Δ < 0), indicating no real roots for the quadratic equation 4x² - 8x + 9 = 0. The coefficient a = 4 is positive, so the parabola opens upwards. Since the parabola does not intersect the x-axis and opens upwards, the quadratic expression is always positive for all real values of x. Therefore, the solution to the inequality 4x² - 8x + 9 > 0 is the set of all real numbers, represented as (-∞, ∞). This result aligns with the pattern that an upward-opening parabola with no real roots yields a quadratic expression that is always positive.
7) Solving -3.6x² - 7.2x < 0
Lastly, we consider the inequality -3.6x² - 7.2x < 0. Here, we can first factor out -3.6x from the expression:
-3.6x(x + 2) < 0.
To make it easier to work with, we can divide both sides by -3.6. Remember that dividing by a negative number reverses the inequality sign:
x(x + 2) > 0.
Now, we find the critical points by setting x(x + 2) = 0. This gives us x = 0 and x = -2. These points divide the number line into three intervals: (-∞, -2), (-2, 0), and (0, ∞). We will test a value from each interval to determine where the inequality holds:
- Interval (-∞, -2): Choose x = -3. Then (-3)(-3 + 2) = (-3)(-1) = 3 > 0. The inequality holds.
- Interval (-2, 0): Choose x = -1. Then (-1)(-1 + 2) = (-1)(1) = -1 < 0. The inequality does not hold.
- Interval (0, ∞): Choose x = 1. Then (1)(1 + 2) = (1)(3) = 3 > 0. The inequality holds.
Thus, the solution to the inequality x(x + 2) > 0 is x < -2 or x > 0. In interval notation, this is (-∞, -2) ∪ (0, ∞). This example showcases a different approach, where factoring and analyzing intervals help solve the inequality.
In conclusion, solving quadratic inequalities involves a combination of algebraic techniques and graphical interpretations. By calculating the discriminant, we can determine whether the quadratic equation has real roots, which helps us understand if the parabola intersects the x-axis. The sign of the coefficient a tells us whether the parabola opens upwards or downwards, further guiding our solution. In cases where the discriminant is negative, and a is positive, the quadratic expression is always positive, leading to solutions of either all real numbers or the empty set depending on the inequality sign. Factoring, as demonstrated in the final example, provides another effective method for solving quadratic inequalities. Mastering these approaches enables a comprehensive understanding and ability to solve a wide range of quadratic inequalities. This skill is vital not only in mathematics but also in various applications in science, engineering, and economics, where optimization and understanding the range of solutions are essential.