Solving Polynomial Inequality $x^2 - 9x + 14 > 0$ And Graphing Solution

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To solve the polynomial inequality x2βˆ’9x+14>0x^2 - 9x + 14 > 0, we need to find the values of xx that satisfy this condition. This involves several steps, including factoring the quadratic expression, finding the critical values, testing intervals, and expressing the solution set in interval notation. Graphing the solution set on a real number line provides a visual representation of the solution.

1. Factoring the Quadratic Expression

First, we factor the quadratic expression x2βˆ’9x+14x^2 - 9x + 14. We are looking for two numbers that multiply to 14 and add up to -9. These numbers are -2 and -7. Therefore, the quadratic expression can be factored as follows:

x2βˆ’9x+14=(xβˆ’2)(xβˆ’7)x^2 - 9x + 14 = (x - 2)(x - 7)

Now, the inequality becomes:

(xβˆ’2)(xβˆ’7)>0(x - 2)(x - 7) > 0

Understanding this factoring step is crucial. By rewriting the quadratic expression in factored form, we can easily identify the critical values where the expression equals zero. Factoring transforms a complex polynomial into a product of simpler terms, making it easier to analyze its sign in different intervals. For instance, when xx is between 2 and 7, one factor is negative and the other is negative, making the product positive. Outside this interval, the signs change, influencing the inequality's outcome. This initial transformation is a cornerstone in solving polynomial inequalities, providing a clear path to finding the solution set. The correct factorization is essential as it directly impacts the subsequent steps in determining the critical values and intervals to test. Therefore, mastering factoring techniques is vital for solving polynomial inequalities accurately and efficiently.

2. Finding the Critical Values

The critical values are the values of xx that make the expression equal to zero. These values are found by setting each factor equal to zero and solving for xx:

xβˆ’2=0ightarrowx=2x - 2 = 0 ightarrow x = 2

xβˆ’7=0ightarrowx=7x - 7 = 0 ightarrow x = 7

The critical values are x=2x = 2 and x=7x = 7. These values divide the real number line into three intervals: (βˆ’ext∞,2)(- ext{∞}, 2), (2,7)(2, 7), and (7,ext∞)(7, ext{∞}).

Identifying these critical values is a pivotal step in solving polynomial inequalities. These points are where the polynomial expression changes its signβ€”from positive to negative or vice versa. In our example, x=2x = 2 and x=7x = 7 serve as these turning points. To find them, we set each factor of the polynomial to zero, as these are the xx values that make the entire expression zero. This process essentially maps out the boundaries of intervals where the inequality’s sign remains consistent. Grasping the significance of critical values is essential as they form the basis for the subsequent interval testing. By determining these values accurately, we can systematically analyze how the polynomial behaves across different sections of the number line. The act of finding critical values isn't just about calculation; it’s about understanding the polynomial’s structure and behavior. This understanding allows us to predict how the expression’s sign will vary, which is crucial for solving the inequality.

3. Testing the Intervals

We need to test a value from each interval to determine whether the inequality (xβˆ’2)(xβˆ’7)>0(x - 2)(x - 7) > 0 is satisfied. We'll pick test values and plug them into the factored inequality:

  1. Interval (βˆ’ext∞,2)(- ext{∞}, 2): Choose x=0x = 0

    (0βˆ’2)(0βˆ’7)=(βˆ’2)(βˆ’7)=14>0(0 - 2)(0 - 7) = (-2)(-7) = 14 > 0 (True)

  2. Interval (2,7)(2, 7): Choose x=4x = 4

    (4βˆ’2)(4βˆ’7)=(2)(βˆ’3)=βˆ’6less0(4 - 2)(4 - 7) = (2)(-3) = -6 less 0 (False)

  3. Interval (7,ext∞)(7, ext{∞}): Choose x=8x = 8

    (8βˆ’2)(8βˆ’7)=(6)(1)=6>0(8 - 2)(8 - 7) = (6)(1) = 6 > 0 (True)

The testing of intervals is where we determine the sign of the polynomial expression across different segments of the number line. By selecting a test value within each intervalβ€”created by the critical valuesβ€”we substitute it into the factored inequality. This substitution reveals whether the expression is positive or negative in that interval. For instance, if a test value from (βˆ’ext∞,2)(- ext{∞}, 2) makes the expression positive, then the entire interval satisfies the inequality if it demands the expression be greater than zero. This step is essential because polynomials can change signs only at their roots (critical values). Interval testing simplifies the task of solving inequalities by converting it into a series of manageable checks. The process effectively narrows down the regions where the inequality holds, eliminating guesswork and ensuring a systematic approach. Accurate testing is vital as errors here will lead to an incorrect solution set. It's a practical application of mathematical principles, turning an abstract problem into a series of concrete evaluations.

4. Expressing the Solution Set in Interval Notation

The intervals where the inequality is true are (βˆ’ext∞,2)(- ext{∞}, 2) and (7,ext∞)(7, ext{∞}). Since the inequality is strict (greater than, not greater than or equal to), we use parentheses to exclude the critical values from the solution set.

The solution set in interval notation is:

(βˆ’ext∞,2)extβˆͺ(7,ext∞)(- ext{∞}, 2) ext{βˆͺ} (7, ext{∞})

Expressing the solution set in interval notation is the final step in clearly defining all xx values that satisfy the original inequality. Interval notation is a concise way to represent continuous ranges of numbers. Parentheses are used to exclude endpoints, indicating that the critical values themselves are not included in the solution, while brackets would include them. The union symbol β€œβˆͺ” combines separate intervals into a single solution set. This notation is not just a formality; it ensures the solution is communicated accurately and universally. It eliminates ambiguity by precisely specifying the boundaries of the solution. The skill of translating the solution from the number line to interval notation is crucial for mathematical communication. It demonstrates a complete understanding of the inequality, from its initial form to its final solution. This representation is essential for further mathematical work, such as graphing or combining solutions with other inequalities.

5. Graphing the Solution Set on a Real Number Line

To graph the solution set on a real number line, we mark the critical values 22 and 77 with open circles (since they are not included in the solution) and shade the intervals (βˆ’ext∞,2)(- ext{∞}, 2) and (7,ext∞)(7, ext{∞}).

The graph looks like this:

<-----------------(    )---------------------(    )-------------------->
              2                          7

This visual representation helps to understand the range of xx values that satisfy the inequality. The open circles at 2 and 7 signify that these points are not included in the solution set, consistent with the strict inequality x2βˆ’9x+14>0x^2 - 9x + 14 > 0. The shaded regions to the left of 2 and to the right of 7 visually confirm the solution intervals (βˆ’ext∞,2)(- ext{∞}, 2) and (7,ext∞)(7, ext{∞}).

Creating a graph of the solution set on a number line provides an invaluable visual aid in understanding the solution to the inequality. The number line graphically represents all real numbers, and by shading the intervals that satisfy the inequality, we create a clear picture of the solution range. Open circles at the critical values (2 and 7 in this case) indicate that these points are not part of the solution, a crucial detail for strict inequalities. This visual representation can immediately convey the solution in a way that algebraic notation might not. For students, the graph reinforces the connection between the algebraic solution and its numerical representation. It’s also a practical tool for checking the solution; a quick glance can confirm whether the shaded regions align with the algebraic findings. Moreover, the graph is essential for more complex problems involving multiple inequalities, where visualizing the overlap (or lack thereof) of solution sets becomes critical. The number line graph, therefore, is not just a visual flourish; it’s an integral component of solving and understanding inequalities.

Final Answer

The solution set of the inequality x2βˆ’9x+14>0x^2 - 9x + 14 > 0 is (βˆ’ext∞,2)extβˆͺ(7,ext∞)(- ext{∞}, 2) ext{βˆͺ} (7, ext{∞}).

In Conclusion, solving polynomial inequalities like x2βˆ’9x+14>0x^2 - 9x + 14 > 0 involves a systematic approach that combines algebraic techniques with visual representation. By factoring the polynomial, identifying critical values, testing intervals, expressing the solution set in interval notation, and graphing the solution on a number line, we can accurately determine the range of xx values that satisfy the inequality. Each step is crucial, building upon the previous one to ensure a comprehensive and correct solution. This method not only solves the specific inequality but also equips us with a robust framework for tackling more complex polynomial inequalities. Understanding these steps and their underlying principles is essential for mastering algebra and its applications in various fields.